3.example

Example 5.8

求矩阵的转置。

输入数据：

2 1 3
3 3 1
1 2 1

输出数据：

2 3 1
1 3 2
3 1 1

[+]

#include <iostream>
#include <iomanip>
using namespace std;
const int n=3;
int a[n+1][n+1];

int main(){

for(int i=1;i<=n;i++){
  for(int j=1;j<=n;j++){
    cin >> a[i][j];
  }
}
for(int i=1;i<=n;i++){
  for(int j=1;j<=n;j++){
    cout << setw(5) << a[j][i];
  }
  cout << endl;
}

return 0;
}

Example 5.9

已知一个6*6的矩阵（方阵），把矩阵两条对角线上的元素加上10，然后输出这个新矩阵。

[+]

#include <iostream>
#include <iomanip>
using namespace std;
int a[7][7];

int main(){
for(int i=1;i<=6;i++)
  for(int j=1;j<=6;j++)
    cin >> a[i][j];
for(int i=1;i<=6;i++)
  for(int j=1;j<=6;j++)
    if((i==j)||(i+j==7))
      a[i][j]+=10;
for(int i=1;i<=6;i++){
  for(int j=1;j<=6;j++){
    cout << setw(5) << a[i][j];
  }
  cout << endl;
}	

return 0;
}

Example 5.10

大部分元素是0的矩阵称为稀疏矩阵，假设有k个非0元素，则可把稀疏矩阵用k*3的矩阵简记之，其中第一列是行号，第二列是列号，第三列是该行、该列下的该元素的值。如：

0 0 0 5            简记成：  1 4 5     //第1行第4列有个数是5
0 2 0 0                     2 2 2     //第2行第2列有个数是2
0 1 0 0                     3 2 1     //第3行第2列有个数是1

试编程读入一个稀疏矩阵，转换成简记形式，并输出。

[+]

#include <iostream>
#include <iomanip>
using namespace std;
const int n=3,m=5;
int a[n+1][m+1],b[101][4],k=0;

int main(){
for(int i=1;i<=n;i++){
  for(int j=1;j<=m;j++)
    cin >> a[i][j];
}
for(int i=1;i<=n;i++)
  for(int j=1;j<=m;j++)
    if(a[i][j]!=0){
      k++;
      b[k][1]=i;
      b[k][2]=j;
      b[k][3]=a[i][j];				
    }

for(int i=1;i<=k;i++){
  for(int j=1;j<=3;j++)
    cout << setw(3) << b[i][j];
  cout << endl;
}

return 0;
}

Example 5.11

打印杨辉三角的前10行。

    1              1
   1 1             1 1
  1 2 1            1 2 1
 1 3 3 1           1 3 3 1
1 4 6 4 1          1 4 6 4 1

[+]

#include <iostream>
#include <iomanip>
using namespace std;

int a[100][100];

int main(){
a[1][1]=1;
for(int i=2;i<=10;i++){
  a[i][1]=1;
  a[i][i]=1;
  for(int j=2;j<i;j++)
    a[i][j]=a[i-1][j-1] + a[i-1][j];
}
for(int i=1;i<=10;i++){
  if(i!=10)
    cout << setw(30-3*i) << " ";
  for(int j=1;j<=i;j++)
    cout << setw(6) << a[i][j];
  cout << endl; 
}

return 0;
}

Example 5.12

输入一串字符，字符个数不超过100，且以“.”结束。判断它们是否构成回文。例如：12321，ABCBA，AA等。

[+]

#include <iostream>
using namespace std;

char ch,letter[101];
int i=0,j=1;

int main(){
cout << "Input a string:";
cin >> ch;
while(ch!='.'){
  letter[++i]=ch;
  cin >> ch;
}
while(j<i && letter[j]==letter[i]){
  i--;
  j++;
}
if(j>=i)
  cout << "Yes" << endl;
else
  cout << "No" << endl;
  
return 0;
}

Example 5.13

蛇形填数：在n*n方阵里填入1，2，3，...，n*n，要求填成蛇形。例如n=4时方阵为：

[+]

#include <iostream>
#include <cstring>
using namespace std;
#define MAXN 100
int a[MAXN][MAXN];
int n,x,y,tot;

int main(){
scanf("%d",&n);
memset(a,0,sizeof(a));
tot=a[x=0][y=n-1]=1;
while(tot<n*n){
  while(x+1<n && !a[x+1][y]) 
    a[++x][y]=++tot;
  while(y-1>=0 && !a[x][y-1])
    a[x][--y]=++tot;
  while(x-1>=0&& !a[x-1][y])
    a[--x][y]=++tot;
  while(y+1<n && !a[x][y+1])
    a[x][++y]=++tot;
}
for(int i=0;i<n;i++){
  for(int j=0;j<n;j++)
    printf("%6d",a[i][j]);
  printf("\n");
}

return 0;
}

Exercise 1

矩阵交换行，http://noi.openjudge.cn/ch0108/01/[+]

#include <iostream>
using namespace std;

int main(){
    int mat[5][5],n,m;
    for(int i=0;i<5;i++){
        for(int j=0;j<5;j++){
            cin >> mat[i][j]; 
        }
    }
    cin >> n >> m;
    n--,m--;
    for(int i=0;i<5;i++){
        int tmp = mat[n][i];
        mat[n][i] = mat[m][i];
        mat[m][i] = tmp;
    }
    
    for(int i=0;i<5;i++){
        cout << mat[i][0] ;
        for(int j=1;j<5;j++){
            cout << " " << mat[i][j];
        }
        cout << endl;
    }
    
    return 0;
}

同行列对角线的格子，http://noi.openjudge.cn/ch0108/02/[+]

#include <iostream>
using namespace std;

int main(){
  int n,x,y;
  cin >> n >> x >> y;
  
    cout << "(" << x << "," << 1 << ")";
  for(int i=2;i<=n;i++)
      cout << " (" << x << "," << i << ")";
  cout << endl;
  
    cout << "(" << 1 << "," << y << ")";
  for(int i=2;i<=n;i++)
      cout << " (" << i << "," << y << ")";
    cout << endl;
  
  int i=x,j=y;
  while(i>1 && j>1){
    i--;
    j--;
  }
  
    cout << "(" << i++ << "," << j++ << ")";
  while(i<=n && j<=n)
      cout << " (" << i++ << "," << j++ << ")";
  cout << endl;

  i=x,j=y;
  while(i<n && j>1){
    i++;
    j--;
  }
    cout << "(" << i-- << "," << j++ << ")";
  while(i>=1 && j<=n)
      cout << " (" << i-- << "," << j++ << ")";
  cout << endl;

    return 0;
}

计算矩阵边缘元素之和，http://noi.openjudge.cn/ch0108/03/[+]

#include <iostream>
using namespace std;

int a[105][105],n,m;
long long sum=0;
int main(){
  cin >> n >> m;
  
  for(int i=0;i<n;i++)
    for(int j=0;j<m;j++)
      cin >> a[i][j];
  
  for(int i=0;i<m;i++)
    sum = sum + a[0][i];
  if(n>1)
    for(int i=0;i<m;i++)
      sum = sum + a[n-1][i];
    
  if(n>2){
    for(int i=1;i<n-1;i++)
      sum = sum + a[i][0];
    if(m>1)
      for(int i=1;i<n-1;i++)
        sum = sum + a[i][m-1];
      
  }
    
  cout << sum << endl;
  
  return 0;
}

错误探测，http://noi.openjudge.cn/ch0108/04/[+]

#include <iostream>
using namespace std;

int a[100][100],n;
int row[100],col[100];
int rn=0,cn=0,r,c;
int main(){
  cin >> n;
  for(int i=0;i<n;i++)
    for(int j=0;j<n;j++){
      cin >> a[i][j];
      if(a[i][j]==1){
        row[i]++;
        col[j]++;
      }
    }	
  for(int i=0;i<n;i++){
    if(row[i]%2!=0){
      rn++;
      r=i+1;
    }
    if(col[i]%2!=0){
      cn++;
      c=i+1;
    } 
  }
  
  if(rn==0 && cn==0)
    cout << "OK" << endl;
  else if(rn==1 && cn==1)
    cout << r << " " << c << endl;
  else
    cout << "Corrupt" << endl;
  
  return 0;
}

计算鞍点，http://noi.openjudge.cn/ch0108/05/[+]

#include <iostream>
using namespace std;

int a[6][6];
int row[6],col[6];
int main(){
  for(int i=0;i<6;i++)
    col[i]=0x7fffffff;
  for(int i=0;i<5;i++)
    for(int j=0;j<5;j++){
      cin >> a[i][j];
      if(a[i][j]>row[i])row[i]=a[i][j];
      if(a[i][j]<col[j])col[j]=a[i][j];
    }
  int flag = 1;
  for(int i=0;i<5;i++)
    for(int j=0;j<5;j++){
      if(row[i]==col[j]){
        cout << i+1 << " " << j+1 << " " << a[i][j] << endl;
        flag = 0; 
      }
    }	
  if(flag)
    cout << "not found" << endl;
  
  return 0;
}

图像相似度，http://noi.openjudge.cn/ch0108/06/[+]

#include <iostream>
using namespace std;

int a[105][105],n,m,p;
double sum = 0;

int main(){
  cin >> n >> m;

  for(int i=0;i<n;i++)
    for(int j=0;j<m;j++)
      cin >> a[i][j];
  for(int i=0;i<n;i++)
    for(int j=0;j<m;j++){
      cin >> p;
      if(p==a[i][j]){
        sum = sum + 1;
      }
    }
        
  printf("%.2f\n",100*sum/(n*m));
  
  return 0;
}

矩阵归零消减序列和，http://noi.openjudge.cn/ch0108/07/[+]

#include <iostream>
using namespace std;

int a[105][105],n;
int row[105],col[105];

void print(){
  for(int i=0;i<n;i++){
    for(int j=0;j<n;j++){
      cout << a[i][j] << " ";
    }
    cout << endl;
  }
  cout << endl;
}

int main(){
  
  cin >> n;
  
  for(int i=0;i<n;i++){
    row[i]=0x7fffffff;
    col[i]=0x7fffffff;
  }
  for(int i=0;i<n;i++){
    for(int j=0;j<n;j++){
      cin >> a[i][j];
      if(a[i][j]<row[i]) row[i]=a[i][j];
    }
  }
  int t = n;
  do{
    cout << a[1][1] << endl;
    
    for(int i=0;i<t;i++){
      for(int j=0;j<t;j++){
        a[i][j] = a[i][j]-row[i];
        if(a[i][j]<col[j]) col[j]=a[i][j];
      }
    }
    //print();
    for(int i=0;i<t;i++){
      for(int j=0;j<t;j++){
        a[j][i] = a[j][i] - col[i];
      }
    }
    //print();
    for(int i=0;i<t;i++){
      row[i]=a[i][0];
      for(int j=1;j<t-1;j++){
        a[i][j]=a[i][j+1];
        if(a[i][j]<row[i]) row[i]=a[i][j];
      }
      col[i]=a[0][i];
    }
    //print();
    for(int i=1;i<t-1;i++){
      for(int j=0;j<t;j++){
        a[i][j]=a[i+1][j];
      }
      row[i]=row[i+1];
    }
  }while(--t>0);
  
  return 0;
}

矩阵加法，http://noi.openjudge.cn/ch0108/08/[+]

#include <iostream>
#include <cstdio>
using namespace std;
int a[105][105];
int n,m,b;
int main(){
  
  cin >> n >> m;
  
  for(int i=1;i<=n;i++)
    for(int j=1;j<=m ;j++)
      cin >> a[i][j];
  
  for(int i=1;i<=n;i++)
    for(int j=1;j<=m;j++){
      cin >> b;
      a[i][j]+=b;
    }		
  
  for(int i=1;i<=n;i++){
    cout << a[i][1];
    for(int j=2;j<=m;j++){
      cout << " " << a[i][j];
    }
    cout << endl;
  }
  
  
  return 0;
}

矩阵乘法，http://noi.openjudge.cn/ch0108/09/[+]

#include <iostream>
#include <cstdio>

using namespace std;
int n,m,k;
int a[105][105],b[105][105],c[105][105];

int main(){
  
  cin >> n >> m >> k;
  
  for(int i=1;i<=n;i++)
    for(int j=1;j<=m;j++)
      cin >> a[i][j];
  
  for(int i=1;i<=m;i++)
    for(int j=1;j<=k;j++)
      cin >> b[i][j];
      
  for(int i=1;i<=n;i++)
    for(int j=1;j<=k;j++)
      for(int l=1;l<=m;l++)
        c[i][j] = c[i][j] + a[i][l] * b[l][j];
        
  for(int i=1;i<=n;i++){
    cout << c[i][1];
    for(int j=2;j<=k;j++){
      cout << " " << c[i][j];
    }
    cout << endl;
  }
      
  
      
  return 0;
}

矩阵转置，http://noi.openjudge.cn/ch0108/10/[+]

#include <iostream>
using namespace std;

int n,m,a[105][105];

int main(){
  cin >> n >> m;
  
  for(int i=0;i<n;i++)
    for(int j=0;j<m;j++)
      cin >> a[i][j];
  
  for(int i=0;i<m;i++){
    for(int j=0;j<n;j++)
      cout << a[j][i] << " ";
    cout << endl;
  }
    
  
  return 0;
}

图像旋转，http://noi.openjudge.cn/ch0108/11/[+]

#include <iostream>
using namespace std;
int n,m,a[105][105];

int main(){
  cin >> n >> m;
  for(int i=0;i<n;i++)
    for(int j=0;j<m;j++)
      cin >> a[i][j];
  
  for(int i=0;i<m;i++){
    for(int j=n-1;j>=0;j--)
      cout << a[j][i] << " ";
    cout << endl;
  }	
  
  return 0;
}

变幻的矩阵，http://noi.openjudge.cn/ch0108/12/[+]

#include <iostream>
using namespace std;

int n;
char a[10][10],b[10][10];
bool judge1();
bool judge2();
bool judge3();
bool judge4();

int main(){
  cin >> n;
  
  for(int i=0;i<n;i++)
    for(int j=0;j<n;j++)
      cin >> a[i][j];
  for(int i=0;i<n;i++)
    for(int j=0;j<n;j++)
      cin >> b[i][j];
  
  if(judge1()) cout << 1 << endl;
  else if(judge2()) cout << 2 << endl;
  else if(judge3()) cout << 3 << endl;
  else if(judge4()) cout << 4 << endl;
  else 	cout << 5 << endl;
  
  return 0;
}

bool judge1(){
  bool ret = true;
  for(int i=0;i<n;i++)
    for(int j=0;j<n;j++)
      if(a[i][j]!=b[j][n-i-1]){
        ret = false;
      }
  return ret;
}

bool judge2(){
  bool ret = true;
  for(int i=0;i<n;i++)
    for(int j=0;j<n;j++)
      if(a[i][j]!=b[n-j-1][i]){
        ret = false;
      }
  return ret;
}

bool judge3(){
  bool ret = true;
  for(int i=0;i<n;i++)
    for(int j=0;j<n;j++)
      if(a[i][j]!=b[n-i-1][n-j-1]){
        ret = false;
      }
  return ret;
}

bool judge4(){
  bool ret = true;
  for(int i=0;i<n;i++)
    for(int j=0;j<n;j++)
      if(a[i][j]!=b[i][j]){
        ret = false;
      }
  return ret;
}

图像模糊处理，http://noi.openjudge.cn/ch0108/13/[+]

#include <iostream>
using namespace std;

int n,m,b[105][105];
double a[105][105];

int main(){
  cin >> n >> m;
  for(int i=0;i<n;i++)
    for(int j=0;j<m;j++){
      cin >> a[i][j];
      b[i][j] = a[i][j];
    }
  if(n>=3 && m>=3)
    for(int i=1;i<n-1;i++)
      for(int j=1;j<m-1;j++){
        b[i][j] = (a[i][j]+a[i][j-1]+a[i][j+1]+a[i-1][j]+a[i+1][j])/5+0.5;
      }
  
  for(int i=0;i<n;i++){
    for(int j=0;j<m;j++)
      cout << b[i][j] << " ";
    cout << endl;
  }
  return 0;
}

扫雷游戏地雷数计算，http://noi.openjudge.cn/ch0108/14/[+]

#include <iostream>
using namespace std;

char d[105][105];
char mine[105][105];
int x[8]={-1,0,1,-1,1,-1,0,1};
int y[8]={-1,-1,-1,0,0,1,1,1};
int n,m;

int main(){
  cin >> n >> m;
  for(int i=0;i<n;i++)
    scanf("%s",d[i]);
    
  for(int i=0;i<n;i++){
    for(int j=0;j<m;j++){
      if(d[i][j]=='*')
        mine[i][j]='*';
      else{
        int tmp = '0';
        for(int k=0;k<8;k++){
          int x1 = i+x[k];
          int y1 = j+y[k];
          if(x1>=0 && x1<=n && y1>=0 && y1<=m){
            if(d[x1][y1]=='*')
              tmp = tmp + 1;
          }
        }
        mine[i][j]=tmp;
      }			
    }
  }
  
  for(int i=0;i<n;i++){
    for(int j=0;j<m;j++){
      cout << mine[i][j];
    }
    cout << endl;
  }
  
  return 0;
}

细菌的繁殖与扩散，http://noi.openjudge.cn/ch0108/15/[+]

#include <iostream>
using namespace std;

long long a[12][12],b[12][12],m,n;
int x[8]={-1,0,1,-1,1,-1,0,1};
int y[8]={-1,-1,-1,0,0,1,1,1};
int main(){
  cin >> m >> n;
  
  a[5][5] = m;
  
  for(int t=1;t<=n;t++){
    for(int i=1;i<=9;i++){
      for(int j=1;j<=9;j++){
        b[i][j] = b[i][j]+ a[i][j]*2;
        for(int k=0;k<8;k++){
          int x1=i+x[k];
          int y1=j+y[k];
          b[x1][y1] = b[x1][y1] + a[i][j];
        }				
      }
    }
    
    for(int i=1;i<=9;i++)
      for(int j=1;j<=9;j++){
        a[i][j] = b[i][j];
        b[i][j] = 0;
      }
  }
  
  for(int i=1;i<=9;i++){
    cout << a[i][1];
    for(int j=2;j<=9;j++)
      cout << " " << a[i][j];
    cout << endl;
  }
  
  return 0;
}

矩阵剪刀石头布，http://noi.openjudge.cn/ch0108/16/[+]

#include <iostream>
using namespace std;

char a[101][101],b[101][101];
int c,r,n;

int main(){
  cin >> r >> c >> n;
  for(int i=0;i<r;i++)
    for(int j=0;j<c;j++){
      cin >> a[i][j];
      b[i][j] = a[i][j];			
    }

  while(n--){
    for(int i=0;i<r;i++){
      for(int j=0;j<c;j++){
        switch(a[i][j]){
          case 'R':
            if((i+1<r&&a[i+1][j]=='P')||(j+1<c&&a[i][j+1]=='P')||(i-1>=0&&a[i-1][j]=='P')||(j-1>=0&&a[i][j-1]=='P'))
              b[i][j]='P';
            break;
          case 'S':
            if((i+1<r&&a[i+1][j]=='R')||(j+1<c&&a[i][j+1]=='R')||(i-1>=0&&a[i-1][j]=='R')||(j-1>=0&&a[i][j-1]=='R'))
              b[i][j]='R';
            break;
          case 'P': 
            if((i+1<r&&a[i+1][j]=='S')||(j+1<c&&a[i][j+1]=='S')||(i-1>=0&&a[i-1][j]=='S')||(j-1>=0&&a[i][j-1]=='S'))
              b[i][j]='S';
            break;
        }
      }
    }
    for(int i=0;i<r;i++)
      for(int j=0;j<c;j++){
        a[i][j] = b[i][j];			
      }
  }
  
  for(int i=0;i<r;i++){
    for(int j=0;j<c;j++){
      cout << a[i][j];			
    }
    cout << endl; 
  }

  return 0;
}

最好的草，http://noi.openjudge.cn/ch0108/17/[+]

#include <iostream>
using namespace std;

char f[101][101];
int r,c;

int main(){
  cin >> r >> c;
  for(int i=0;i<r;i++)
    cin >> f[i];
  int cnt = 0;
  for(int i=0;i<r;i++){
    for(int j=0;j<c;j++){
      if(f[i][j]=='#'){
        f[i][j]='.';
        if(f[i][j+1]=='#'){
          f[i][j+1]='.';
        }else if(f[i+1][j]=='#'){
          f[i+1][j]='.';
        }
        cnt++;
      }
    }
  }
  cout << cnt << endl;
  
  return 0;
}

肿瘤面积，http://noi.openjudge.cn/ch0108/18/[+]

#include <iostream>
using namespace std;

int p[1005][1005],n,x1,y1,x2,y2;

int main(){
  cin >> n;
  
  for(int i=0;i<n;i++)
    for(int j=0;j<n;j++)
      cin >> p[i][j];

  for(int i=0;i<n;i++)
    for(int j=0;j<n;j++){
      if(p[i][j]==0){
        if(i<(n-1) && j<(n-1) && p[i+1][j]==0&&p[i][j+1]==0){
          x1=i;
          y1=j;
        }else if(j>0&&i>0&&p[i-1][j]==0&&p[i][j-1]==0){
          x2=i;
          y2=j;
        }
      }
    }

  if(x1!=0 || x2!=0)
    cout << (x2-x1-1)*(y2-y1-1) << endl;
  else
    cout << 0 << endl;
    
  return 0;
}

肿瘤检测，http://noi.openjudge.cn/ch0108/19/[+]

#include <iostream>
using namespace std;

int p[105][105],n,s,l;

int main(){
  cin >> n;
  for(int i=0;i<n;i++){
    for(int j=0;j<n;j++){
      cin >> p[i][j];
    }
  }
  for(int i=0;i<n;i++){
    for(int j=0;j<n;j++){
      if(p[i][j]<=50){
        s++;
        if(i==0 || j==0 || i==n-1 || j==n-1)
          l++;
        else if(p[i-1][j]>50 || p[i+1][j]>50 || p[i][j-1]>50 || p[i][j+1]>50)
          l++;
      } 
    }
  }
  
  cout << s << " " << l << endl;
  
  return 0;
}

反反复复，http://noi.openjudge.cn/ch0108/20/[+]

#include <iostream>
#include <cstring>
using namespace std;

char w[105][21],s[205];
int n,x,y;

int main(){
  cin >> n >> s;
  
  int len = strlen(s);
  x = -1;	
  for(int i=0;i<len;i++){
    y = i %n;
    if(y==0) x++;
    if(x%2==0)
      w[x][y] = s[i];
    else
      w[x][n-y-1] = s[i];
  }
  
  for(int j=0;j<n;j++){
    for(int i=0;i<=x;i++){
      cout << w[i][j];
    }
  }
  cout << endl;
  
  return 0;
}

二维数组右上左下遍历，http://noi.openjudge.cn/ch0108/21/[+]

#include <iostream>
using namespace std;

int row,col,arr[105][105],x,y,tot,r,c;

int main(){
  cin >> row >> col;
  for(int i=0;i<row;i++){
    for(int j=0;j<col;j++){
      cin >> arr[i][j];
    }
  }
  tot = row * col;
  r=x = 0;
  c=y = 0;
  while(tot--){
    cout << arr[x][y] << endl;
    if(y==0 || x==row-1){
      c++;
      if(c>=col){
        y = col-1;
        r++;
        x = r;
      }else{
        y=c;
        x=0;
      }
    }else{
      x++;
      y--;			
    }
  }

  return 0;
}

神奇的幻方，http://noi.openjudge.cn/ch0108/22/[+]

#include <iostream>
using namespace std;

int n,m[50][50],x1,x2,y1,y2,c,t,p;

int main(){
  cin >> n;
  x1 = 0;
  y1 = n-1;
  p = 1;
  t=(2*n-1)*(2*n-1)-1;
  m[x1][y1] = p;
  
  while(t--){
    if(x1==0){
      if(y1==2*n-2){
        x2 = x1+1;
        y2 = y1;
      }else{
        x2 = 2*n-2;
        y2= y1+1;
      }
    }else if(y1==2*n-2){
      y2=0;
      x2=x1-1;
    }else{
      y2=y1+1;
      x2=x1-1;
      if(m[x2][y2]!=0){
        x2=x1+1;
        y2=y1;
      }
    }
    m[x2][y2]=++p;
    
    x1 = x2;
    y1 = y2;
  }
  
  t = 2*n-1;
  for(int i=0;i<t;i++){
    cout << m[i][0];
    for(int j=1;j<t;j++){
      cout << " " << m[i][j];
    }
    cout << endl;
  }
  
  return 0;
}

二维数组回形遍历，http://noi.openjudge.cn/ch0108/23/[+]

#include <iostream>
using namespace std;

int a[100][100],n,m,row=1,col=1;
int lft,rgt,top,btm;

int main(){
  cin >> n >> m;
  for(int i=1;i<=n;i++){
    for(int j=1;j<=m;j++){
      cin >> a[i][j];
    }
  }
  
  while(row<=n||col<=m){
    for(int i=row,j=col;j<=m&&row<=n;j++)
      cout << a[i][j] << endl;
    row++;
    for(int i=row,j=m;i<=n&&col<=m;i++)
      cout << a[i][j] << endl;
    m--;
    for(int i=n,j=m;j>=col&&row<=n;j--)
      cout << a[i][j] << endl;
    n--;
    for(int i=n,j=col;i>=row&&col<=m;i--)
      cout << a[i][j] << endl;
    col++;
  }

  return 0;
}

蛇形填充数组，http://noi.openjudge.cn/ch0108/24/[+]

#include <iostream>
#include <cstdlib>
using namespace std;

int a[20][20],x,y,t,n,p=1;

int main(){
  cin >> n;
  t = n*n;
  x=0;
  y=0;
  a[x][y]=p++;
  while(p<=t){
    if(x==0){
      y++;
      a[x][y]=p++;
      while(y!=0 && p<=t){
        x++;
        y--;
        a[x][y]=p++;
      }
    }
    if(y==0 && x!=n-1 && p<=t){
      x++;
      a[x][y]=p++;
      while(x!=0){
        x--;
        y++;
        a[x][y]=p++;
      }
    }
    if(x==n-1 && p<=t){
      y++;
      a[x][y]=p++;
      while(y!=n-1){
        x--;
        y++;
        a[x][y]=p++;
      }
    }
    if(y==n-1 && p<=t){
      x++;
      a[x][y]=p++;
      while(x!=n-1){
        x++;
        y--;
        a[x][y]=p++;
      }
    }
  }

  for(int i=0;i<n;i++){
    cout << a[i][0];
    for(int j=1;j<n;j++){
      cout << " " << a[i][j];
    }
    cout << endl;
  }
  
  return 0;
}

螺旋加密，http://noi.openjudge.cn/ch0108/25/[+]

#include <iostream>
#include <string>
using namespace std;

int r,c;
char m[21][21],dst[21][21];
string s,src;

string getc(char ch){
  char t[6];
  int cn=0;
  if(ch!=' ') cn = ch-'A' + 1;
  for(int i=0;i<5;i++){
    t[4-i]=char(48+cn%2);
    cn= cn /2;
  }
  t[5]='\0';
  return t;
}

int main(){
  cin >> r >> c ;
  int x,y,row=0,col=0;
  x=r;
  y=c;
  getchar();
  getline(cin,s);
  src="";
  for(int i=0;i<s.size();i++)
    src = src + getc(s[i]);
  int tt = r*c,p=0;
  for(int i=src.size();i<=tt;i++)
    src = src + '0';
    
  while(col<=c || row<=r){
    for(int i=row,j=col;j<c&&row<r;j++)
      dst[i][j] = src[p++];
    row++;
    for(int i=row,j=c-1;i<r&&col<c;i++)
      dst[i][j] = src[p++];
    c--;		
    for(int i=r-1,j=c-1;j>=col&&row<r;j--)
      dst[i][j] = src[p++];
    r--;
    for(int i=r-1,j=col;i>=row&&col<c;i--)
      dst[i][j] = src[p++];
    col++;
  }	
  for(int i=0;i<x;i++){
    for(int j=0;j<y;j++)
      cout << dst[i][j];
  }
  
  return 0;
}

Example 5.14

按字母表顺序和逆序每隔一个字母打印，即打印出：

a c e g i k m o q s u w y
z x v t r p n l j h f d b

[+]

#include <iostream>
#include <iomanip>
using namespace std;

int main(){

for(char letter='a';letter<='z';letter+=2)
  cout << setw(3) << letter;
cout << endl;
for(char letter='z';letter>='a';letter-=2)
  cout << setw(3) << letter;
cout << endl;

return 0;
}

Example 5.15

在应用计算机编辑文档的时候，我们经常遇到替换任务。如把文档中的“电脑”都替换成“计算机”。现在请你编程模拟一下这个操作。

输入两行内容，第1行是原文（长度不超过200个字符），第2行包含以空格分隔的两个字符A和B，要求将原文中所有字符A都替换成字符B。注意：区分大小写字母。

输入样例：
I Love China. I love Beijing.
I U

输出样例：
U Love China. U love Beijing.

[+]

#include <iostream>
using namespace std;

char a,b,str[200];
int n=0;

int main(){
while((str[n++]=getchar())!='\n');
a=getchar();
getchar();
b=getchar();

for(int i=0;i<n;i++)
  if(str[i]==a)
    cout << b;
  else
    cout << str[i];
cout << endl;	

return 0;
}

Example 5.16

过滤多余的空格。一个句子中也许有多个连续的空格，过滤掉多余的空格，只留下一个空格。
输入：一行，一个字符串（长度不超过200），句子的头和尾都没有空格。
输出：过滤之后的句子。
样例输入：Hello world. This is c language.
样例输出：Hello world. This is c language.

[+]

#include <iostream>
using namespace std;
char st[200];

int main(){

while(scanf("%s",st)==1) printf("%s ",st);

return 0;
}

Example 5.17

C++中，一个字符串中的字符可以通过其对应的下标灵活使用。

[+]

#include <iostream>
#include <cstring>
using namespace std;

char st[100];

int main(){

gets(st);
for(int i=0;i<strlen(st);i++)
  cout << st[i];
cout << endl;

return 0;
}

Example 5.18

对给定的10个国家名，按其字母的顺序输出。

[+]

#include <iostream>
#include <cstring>
using namespace std;

char t[21],cname[11][12];

int main(){
for(int i=1;i<=10;i++)
  gets(cname[i]);
for(int i=1;i<=10;i++){
  int k=i;
  for(int j=i+1;j<=10;j++)
    if(strcmp(cname[k],cname[j])>0)
      k=j;
  strcpy(t,cname[i]);
  strcpy(cname[i],cname[k]);
  strcpy(cname[k],t);
}
for(int i=1;i<=10;i++)
  cout << cname[i] << endl;

return 0;
}

[+]

#include <iostream>
#include <string>
#include <algorithm>
using namespace std;

string cname[10];

int main(){
for(int i=0;i<10;i++)
  getline(cin,cname[i]);
sort(cname,cname+10);
for(int i=0;i<10;i++)
  cout << cname[i] << endl;
return 0;
}

Example 5.19

字符串判等：判断两个由大小写字母和空格组成的字符串在忽略大小写，且忽略空格后是否相等。
输入：两行，每行包含一个字符串。
输出：若两个字符串相等，输出YES，否则输出NO。
输入样例：

a A bb BB ccc CCC
Aa BBbb CCCccc

输出样例：

YES

[+]

#include <iostream>
#include <cstring>
using namespace std;

const int N = 256;
char s1[N],s2[N],a[N],b[N];
int l1,l2;

int main(){
gets(s1);
gets(s2);
strlwr(s1);
strlwr(s2);
for(int i=0;i<strlen(s1);i++)
  if(s1[i]!=' ')
    a[l1++] = s1[i];
for(int i=0;i<strlen(s2);i++)
  if(s2[i]!=' ')
    b[l2++] = s2[i];
if(strcmp(a,b)==0)
  printf("YES\n");
else
  printf("NO\n");
return 0;
}

Example 5.20

字符串移位包含问题。对于一个字符串来说，定义一次循环移位操作为：将字符串的第一个字符移动到末尾形成新的字符串。

给定两个字符串s1和s2，要求判定其中一个字符串是否是另一个字符串通过若干次循环移位后的新字符串的子串。例如CDAA是由AABCD两次移位后产生的新串BCDAA的子串，而ABCD与ACBD则不能通过多次移位来得到其中一个字符串是新的子串。

输入：一行，包含两个字符串，中间由单个空格隔开。字符串只包含字母和数字，长度不超过30。
输出：如果一个字符串是另一个字符串通过若干次循环移位产生的新串的子串，则输出true，否则输出false。
输入样例：AABCD CDAA
输出样例：true

[+]

#include <iostream>
#include <cstring>
using namespace std;

const int N=61;
char s1[N],s2[N],x[N],t[N];
int l1,l2;

int main(){
scanf("%s%s",&s1,&s2);
if(strlen(s1)<strlen(s2)){
  strcpy(t,s1);
  strcpy(s1,s2);
  strcpy(s2,t);
}
strcpy(x,s1);
if(strstr(strcat(s1,x),s2)==NULL)
  printf("false\n");
else
  printf("true\n");
return 0;
}

Exercise 2

统计数字字符个数，http://noi.openjudge.cn/ch0107/01/[+]

#include <iostream>
#include <cstring>
using namespace std;

int main(){
    char cs[300];
    
    cin.getline(cs,300);
    int len = strlen(cs);
    
    int sum = 0;
    for(int i=0;i<len;i++)
        if(cs[i]>='0' && cs[i]<='9')
            sum = sum + 1;
    cout << sum << endl;
    
    return 0;
}

找第一个只出现一次的字符，http://noi.openjudge.cn/ch0107/02/[+]

/*
example 1:
  abcabcdefabcd
  abcdef
  333211
  abcabcdefabcd
  33333321
  
steps:
step1: read string into memory
step2: count number of each character
step3: find first character of number one
step4: output answer
*/

#include <iostream>
using namespace std;

int count[26];

int main(){
  char txt[100005];
  int ntxt=0,flag=0;

  while((txt[ntxt]=getchar())!='\n'){
    count[txt[ntxt]-'a'] = count[txt[ntxt]-'a']+1;
    ntxt++;
  }

  for(int i=0;i<ntxt;i++)
    if(count[txt[i]-'a']==1){
      cout << txt[i] << endl;
      flag=1;
      break;
    }

  if(!flag){
    cout << "no" << endl;
  }

  return 0;
}

基因相关性，http://noi.openjudge.cn/ch0107/03/[+]

#include <iostream>
#include <cstring>
using namespace std;

int main(){
  double h,count=0.0;
  char a[500];
  char b[500];
  
  cin >> h;
  cin >> a;
  cin >> b;
  
  int len = strlen(a);

  for(int i=0;i<len;i++){
    if(a[i]==b[i]){
      count = count + 1;
    }
  }
  
  if(count/len>h)
    cout << "yes" << endl;
  else
    cout << "no" << endl;
    
  return 0;
}

石头剪子布，http://noi.openjudge.cn/ch0107/04/[+]

#include <iostream>
using namespace std;

int p(char a){
  return (int)a;
}


int main(){
  int n;
  char s1[10],s2[10];
  
  cin >> n;
  for(int i=0;i<n;i++){
    cin >> s1 >> s2 ;	
    if(p(s1[0])==p(s2[0])){
      cout << "Tie" << endl;
    }else{
      if(p(s1[0])==80){ 		//player1==Paper
        if(p(s2[0])==82)		//player2=Rock
          cout << "Player1" << endl;
        else					//player2=Scissors
          cout << "Player2" << endl;
      }else if(p(s1[0])==82){ //player1=Rock
        if(p(s2[0])==80)		//player2=paper
          cout << "Player2" << endl;
        else					//player2=Scissors
          cout << "Player1" << endl;
      }else{ 					//player1=Scissors
        if(p(s2[0])==80)		//player2=paper
          cout << "Player1" << endl;
        else					//player2=rock
          cout << "Player2" << endl;
      }
    }
  }
  
  return 0;
}

输出亲朋字符串，http://noi.openjudge.cn/ch0107/05/[+]

#include <iostream>
using namespace std;

string s;

int main(){
    getline(cin,s);
    for(int i=0;i<s.size()-1;i++){
      cout << char(s[i]+s[i+1]);
  }
  cout << char(s[0]+s[s.size()-1]) << endl;
  
    return 0;
}

合法 C 标识符，http://noi.openjudge.cn/ch0107/06/[+]

#include <iostream>
using namespace std;

int main(){
  string s;
  getline(cin,s);
  int i=0;
  for(;i<s.size();i++){
    if(s[0]>='0' && s[0]<='9'){
      break;
    }
    if(s[i]=='_' || (s[i]>='0' && s[i]<='9') || (s[i]>='a' && s[i]<='z') || (s[i]>='A' && s[i]<='Z'))
          continue;
    else
      break;
        
  }
  if(i==s.size()){
    cout << "yes" << endl;
  }else{
    cout << "no" << endl;
  }
  
    return 0;
}

配对碱基链，http://noi.openjudge.cn/ch0107/07/[+]

#include <iostream>
using namespace std;

int main(){
  string s;
  getline(cin,s);
  for(int i=0;i<s.size();i++){
    switch(s[i]){
      case 'A':cout << 'T';break;
      case 'T':cout << 'A';break;
      case 'G':cout << 'C';break;
      case 'C':cout << 'G';break;
    }
  }
    return 0;
}

字符替换，http://noi.openjudge.cn/ch0107/08/[+]

#include <iostream>
#include <cstring>
using namespace std;

char str[40],src,dst;

int main(){
  
  cin >> str >> src >> dst;
  
  int len = strlen(str);
  
  for(int i=0;i<len;i++){
    if(str[i]==src){
      str[i] = dst;
    }	
  }
  
  for(int i=0;i<len;i++){
    cout << str[i];
  }
  cout << endl;
  
  return 0;
}

密码翻译，http://noi.openjudge.cn/ch0107/09/[+]

#include <iostream>
using namespace std;

int main(){
    string s;

    getline(cin,s);
    for(int i=0;i<s.size();i++){
        if(s[i]>='a' && s[i]<='z'){
            s[i] = 'a' + (s[i]-'a' +1) % 26;
        }
        if(s[i]>='A' && s[i]<='Z'){
            s[i] = 'A' + (s[i]-'A' +1) % 26;
        }
    }
    cout << s << endl;
    
    return 0;
}

简单密码，http://noi.openjudge.cn/ch0107/10/[+]

#include <iostream>
using namespace std;

int main(){
  int tmp;
  string s;
  
  getline(cin,s);
  for(int i=0;i<s.size();i++){
    if(s[i]>='A' && s[i]<='Z'){
      tmp = s[i]-5;
      if(tmp < 65)
          s[i]='Z'+tmp-64;
      else
        s[i] = tmp;
    }
  }
  cout << s << endl;
  
    return 0;
}

潜伏者，http://noi.openjudge.cn/ch0107/11/[+]

#include <iostream>
#include <cstring>
using namespace std;

char inf[105],src[105],dst[105];
int book[300],num[300],inflen,srclen,dstlen;
bool flag=true;

int main(){
  scanf("%s",inf);
  scanf("%s",src);
  scanf("%s",dst);
  inflen = strlen(inf);
  srclen = strlen(src);
  dstlen = strlen(dst);

  for(int i=0;i<inflen;i++){
    if(book[inf[i]]==0 && src[i]!=0){
      book[inf[i]]=src[i];
    }else if(book[inf[i]]!=src[i]){
      flag=false; 
    }
  }
  
  if(flag){
    for(int i=65;i<=90;i++){
      if(book[i]==0){
        flag=false;
        break;
      }
    }
  }
  
  for(int i=0;i<srclen;i++){
    if(num[src[i]]==0 && inf[i]!=0){
      num[src[i]]=inf[i];
    }else if(num[src[i]]!=inf[i]){
      flag = false;
    }
  }

  if(flag){
    for(int i=65;i<=90;i++){
      if(num[i]==0){
        flag=false;
        break;
      }
    }
  }
  
  if(flag){
    for(int i=0;i<dstlen;i++){
      cout << char(book[dst[i]]);
    }
    cout << endl;
  }else{
    cout << "Failed" << endl;
  }
  
  return 0;
}

加密的病历单，http://noi.openjudge.cn/ch0107/12/[+]

#include <iostream>
#include <cstring>
using namespace std;

char src[100],dst[100];

int main(){
  cin >> src;
  
  int len = strlen(src);
  for(int i=0;i<len;i++){
    char c = src[len-i-1];
    if(c>='A' && c<='Z'){
      c = c + 35;
      if(c>'z') c = 'a' + (c-'z'-1);
    } else if(c>='a' && c<='z'){
      c = c - 29;
      if(c>'Z') c = 'A' + (c-'Z'-1);
    }
    dst[i] = c;
  }
  
  cout << dst << endl;
  
  return 0;
}

将字符串中的小写字母转换成大写字母，http://noi.openjudge.cn/ch0107/13/[+]

#include <iostream>
#include <cstring>
using namespace std;

char s[105];

int main(){
  gets(s);
  
  int len = strlen(s);
  for(int i=0;i<len;i++){
    if(s[i]>='a' && s[i]<='z')
      s[i]=s[i]-32;
  }
  cout << s << endl;
  
  return 0;
}

大小写字母互换，http://noi.openjudge.cn/ch0107/14/[+]

#include <iostream>
#include <cstring>
using namespace std;

char s[90];

int main(){

  gets(s);
  int len = strlen(s);
  
  for(int i=0;i<len;i++){
    if(int(s[i])>=65 && int(s[i])<=90)
      cout << char(s[i]+32);
    else if(int(s[i])>=97 && int(s[i])<=122)
      cout << char(s[i]-32);
    else
      cout << s[i];
  }	
  cout << endl;
  
  return 0;
}

整理药名，http://noi.openjudge.cn/ch0107/15/[+]

#include <iostream>
#include <cstring>
using namespace std;

int n,len;
char s[100];

int main(){
  
  cin >> n;
  getchar();
  for(int i=0;i<n;i++){
    gets(s);
    len = strlen(s);
    if(s[0]>='a' && s[0]<='z')
      s[0]=s[0]-32;
    for(int j=1;j<len;j++){
      if(s[j]>='A' && s[j]<='Z')
        s[j]=s[j]+32;
    }
    cout << s << endl;
  }
  
  return 0;
}

忽略大小写的字符串比较，http://noi.openjudge.cn/ch0107/16/[+]

#include <iostream>
#include <cstring>
using namespace std;

char src[100],dst[100];
int ls,ld,i;

void toUp(char * a){
  int len= strlen(a);
  for(int i=0;i<len;i++){
    if(a[i]>='a' && a[i]<='z')
      a[i]=a[i]-32;
  }
}

int main(){
  gets(src);
  toUp(src);
  gets(dst);
  toUp(dst);
  
  ls = strlen(src);
  ld = strlen(dst);	
  int len = ls < ld ? ls:ld;
  for(i=0;i<len;i++){
    if(src[i]>dst[i]){
      cout <<">" << endl;
      break;
    }else if(src[i]<dst[i]){
      cout << "<" << endl;
      break;
    }		
  }
  if(i==len){
    if(ls==ld){
      cout << "=" << endl;
    }else{
      char c = ls<ld?'<':'>'; 
      cout << c << endl;
    }	
  }
  
  return 0;
}

字符串判等，http://noi.openjudge.cn/ch0107/17/[+]

#include <iostream>
#include <string>
using namespace std;

string src1,src2,dst1,dst2;

string filt(string s){
  string tmp = "";
  int len = s.length();
  for(int i=0;i<len;i++){
    if(s[i]!=' ')
      if(s[i]>='a' && s[i]<='z')
        tmp = tmp + char(s[i]-32);
      else
        tmp = tmp + s[i];
  }
  return tmp;
}

int main(){
  getline(cin,src1);
  getline(cin,src2);
  dst1 = filt(src1);
  dst2 = filt(src2);
  
  if(dst1.compare(dst2)==0){
    cout << "YES" << endl;
  }else{
    cout << "NO" << endl;
  }	

  return 0;
}

验证子串，http://noi.openjudge.cn/ch0107/18/[+]

#include <iostream>
#include <string>
using namespace std;

string s1,s2;

int main(){
  getline(cin,s1);
  getline(cin,s2);
  string::size_type idx1,idx2;
  
  idx1 = s1.find(s2);
  idx2 = s2.find(s1);
  if(idx1==string::npos && idx2==string::npos){
    cout << "No substring" << endl;
  }else if(idx1==string::npos){
    cout << s1 <<" is substring of " << s2 << endl;
  }else{
    cout << s2 <<" is substring of " << s1 << endl;
  }

  return 0;
}

字符串移位包含问题，http://noi.openjudge.cn/ch0107/19/[+]

#include <iostream>
#include <string>
using namespace std;

string s1,s2,d1,d2;

int main(){
  cin >> s1 >> s2;
  
  d1 = s1+s1;
  d2 = s2+s2;
  
  if(d1.find(s2)!=-1 && s2.size()<=s1.size()){
    cout << "true" << endl;
  }else if(d2.find(s1)!=-1 && s1.size()<=s2.size()){
    cout << "true" << endl;
  }else{
    cout << "false" << endl;
  }
  
  return 0;
}

删除单词后缀，http://noi.openjudge.cn/ch0107/20/[+]

#include <iostream>
#include <cstring>
using namespace std;

char post[3][5]={"er","ly","ing"},s[50];

int main(){
  cin >> s;
  int l = strlen(s);
  for(int i=0;i<3;i++){
    int len = strlen(post[i]),j;
    for(j=0;j<len;j++){
      if(post[i][len-j-1]!=s[l-j-1])
        break;			
    }
    if(j==len){
      s[l-j] = '\0';
      break;
    }
  }
  cout << s << endl;
  
  return 0;
}

单词替换，http://noi.openjudge.cn/ch0107/21/[+]

#include <iostream>
#include <string>
using namespace std;

string st[100],src,dst;
int n=0;

int main(){
  while(cin >> st[n++]);
  src = st[n-3];
  dst = st[n-2];
  n=n-3;
  for(int i=0;i<n;i++){
    if(st[i].compare(src)==0){
      st[i] = dst;
    }
  }
  
  cout << st[0];
  for(int i=1;i<n;i++)
    cout << " " << st[i];
  cout << endl; 

  return 0;
}

紧急措施，http://noi.openjudge.cn/ch0107/22/[+]

#include <iostream>
#include <string>
using namespace std;

string email,mail,usr,psw;
int n,flag=1;

int main(){
  cin >> email >> n;
  
  for(int i=0;i<n;i++){
    cin >>	usr >> psw >> mail;
    if(email.compare(mail)==0){
      flag = 0;
      int len = psw.size();
      for(int i=0;i<len;i++){
        if(psw[i]>='A' && psw[i]<='Z')
          psw[i] = psw[i] + 32;
        else if(psw[i]>='a' && psw[i]<='z')
          psw[i] = psw[i] - 32;
      }
      cout << usr << " " << psw << endl;
    }
  }
  
  if(flag){
    cout << "empty" << endl;
  }

  return 0;
}

过滤多余的空格，http://noi.openjudge.cn/ch0107/23/[+]

#include <iostream>
#include <string>
using namespace std;

int main(){
    string s,temp;

    cin >> s;
    while(cin >> temp){
        s += ' ' + temp;
    }
    cout << s << endl;
    
    return 0;
}

单词的长度，http://noi.openjudge.cn/ch0107/24/[+]

#include <iostream>
#include <cstring>
using namespace std;

char s[1005];

int main(){
  
  scanf("%s",s);
  cout << strlen(s);
    
  while(scanf("%s",s)==1){
    cout << "," << strlen(s);
  }
  cout << endl;
  
  return 0;
}

最长最短单词，http://noi.openjudge.cn/ch0107/25/[+]

#include <iostream>
#include <string>
using namespace std;
string st[300];
int n=0,mn=100000,mx=0,pmax,pmin;

int main(){
  while(cin>>st[n]){
    int len = st[n].size();
    if(st[n][0]==',') st[n].erase(0,1);
    if(st[n][len-1]==',') st[n].erase(len-1,0);
    n++;
  }

  for(int i=0;i<n;i++){
    if(st[i].length()>mx){
      mx = st[i].length();
      pmax = i;
    }
    if(st[i].length()<mn){
      mn = st[i].length();
      pmin = i;
    }
  }
  
  cout << st[pmax] << endl;
  cout << st[pmin] << endl;
    
  return 0;
}

字符串最大跨距，http://noi.openjudge.cn/ch0107/26/[+]

#include <iostream>
#include <string>
using namespace std;

string s,src,lft,rgt;
int len,l,r;

int main(){
  cin >> s;
  len = s.length();
  l = s.find_first_of(",");
  r = s.find_last_of(",");
  src = s.substr(0,l);
  lft = s.substr(l+1,r-l-1);
  rgt = s.substr(r+1,len-r-1);

  l = src.find(lft);
  r = src.rfind(rgt);
  
  if(l==-1 || r ==-1){
    cout << -1 << endl;
  }else{
    if(r>=l+lft.length())
      cout << r -l -lft.length() << endl;
    else 
      cout << "-1" << endl;
  }

  return 0;
}

单词翻转，http://noi.openjudge.cn/ch0107/27/[+]

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;

char str[505];
int l,r=0;

int main(){
  gets(str);

  int len = strlen(str);
  while(r<len){
    while(str[r]==' '&&r<len) r++;
    l=r;
    while(str[r]!=' ' && r<len){
      r++;
    }
    reverse(str+l,str+r);
  }
  
  cout << str << endl;
  
  return 0;
}

单词倒排，http://noi.openjudge.cn/ch0107/28/[+]

#include <iostream>
#include <string>
#include <vector>
using namespace std;

vector<string> vs;
string s;

int main(){
  while(cin>>s){
    vs.push_back(s);
  }
  
  int len = vs.size();
  cout << vs[len-1];
  for(int i=len-2;i>=0;i--)
    cout << " " << vs[i];
  cout << endl;
  
  return 0;
}

ISBN号码，http://noi.openjudge.cn/ch0107/29/[+]

#include <iostream>
#include <string>
using namespace std;

string isbn;
int sum = 0,p=1;

int main(){
  
  cin >> isbn;
  
  sum = 0;	
  for(int i=0;i<11;i++)
    if(isbn[i]!='-')
      sum = sum + (isbn[i]-'0')*p++;
  
  if((isbn[12]=='X' && sum%11==10) || (isbn[12]-'0'==sum%11))
    cout << "Right" << endl;
  else if(sum%11==10){
    isbn[12]='X';
    cout << isbn <<endl;
  }else{
    isbn[12] = sum%11+'0';
    cout << isbn << endl; 
  }
    
  return 0;
}

字符环，http://noi.openjudge.cn/ch0107/30/[+]

#include <iostream>
#include <string>
using namespace std;

string s1,s2,s;
int len,mx=0;

int main(){
  
  cin >> s1 >> s2;
  if(s1.length()>s2.length()){
    s = s1;
    s1= s2;
    s2 = s;
  }	

  len = s1.length();

  s1 = s1 + s1;
  s2 = s2 + s2;

  mx = 0;
  for(int l=1;l<=len;l++){
    for(int i=0;i<=2*len-l;i++){
      s = s1.substr(i,l);
      if(s2.find(s)!=string::npos){
        if(s.size()>mx){
          mx = s.size();
        }
      }
    }
  }
  
  cout << mx << endl;
  
  return 0;
}

字符串p型编码，http://noi.openjudge.cn/ch0107/31/[+]

#include <iostream>
#include <cstring>
#include <string>
#include <stack>
using namespace std;

char s[1005],ans[2005],d;
int len,cnt,p;

string ntos(int a){
  stack<char> stk;
  string t = "";
  while(a){
    stk.push(char(a%10+48));
    a=a/10;
  }
  while(!stk.empty()){
    t = t + stk.top();
    stk.pop();
  }
  return t;
}

int main(){
  
  cin >> s;
  len = strlen(s);
  d=s[0];
  cnt = 1;
  p=0;
  for(int i=1;i<len;i++){
    if(s[i]!=d){
      string r = ntos(cnt);
      for(int j=0;j<r.size();j++){
        ans[p++] = r[j];
      }
      ans[p++] = d;
      cnt=1;
      d=s[i];
    }else{
      cnt++;
    }
  }
  string r = ntos(cnt);
  for(int j=0;j<r.size();j++){
    ans[p++] = r[j];
  }
  ans[p++] = d;
  
  cout << ans << endl;
  
  return 0;
}

行程长度编码，http://noi.openjudge.cn/ch0107/32/[+]

#include <iostream>
#include <cstring>
#include <string>
#include <stack>
using namespace std;

char s[1005],ans[2005],d;
int len,cnt;

void toup(char a[]){
  int l = strlen(a);
  for(int i=0;i<l;i++)
    if(a[i]>='a' && a[i]<='z')
      a[i]=a[i]-32;
}

int main(){
  
  cin >> s;
  toup(s);
  
  len = strlen(s);
  d=s[0];
  cnt = 1;
  for(int i=1;i<len;i++){
    if(s[i]!=d){
      cout << "(" << d << "," << cnt << ")";
      cnt=1;
      d=s[i];
    }else{
      cnt++;
    }
  }
  cout << "(" << d << "," << cnt << ")" << endl;
  
  return 0;
}

判断字符串是否为回文，http://noi.openjudge.cn/ch0107/33/[+]

#include <iostream>
#include <cstring>
using namespace std;

char c[105];
int len,flag=1,mid;

int main(){
  cin >> c ;
  len = strlen(c);
  mid = len/2;
  for(int i=0;i<mid;i++){
    if(c[i]!=c[len-i-1]){
      flag=0;
      break;
    }
  }
  if(flag)
    cout << "yes" << endl;
  else
    cout << "no" << endl;

  return 0;
}

回文子串，http://noi.openjudge.cn/ch0107/34/[+]

#include <iostream>
#include <string>
using namespace std;

string s;

int main(){
  cin >> s;
  int len = s.size(),end,right,left;
  
  for(int i=2;i<=len;i++){ //枚举子串的长度 
    end = len - i;
    for(int begin=0;begin<=end;begin++){
      right = begin+i-1;
      for(left=begin;left<right;left++,right--){
        if(s[left]!=s[right])
          break;
      }
      if(left>=right){
        right = begin+i-1;
        for(left=begin;left<=right;left++)
          cout << s[left];
        cout << endl;
      }
    }
    
  }	
  return 0;
}

字符串的展开，http://noi.openjudge.cn/ch0107/35/[+]

#include <iostream>
#include <string>
using namespace std;

int p1,p2,p3;
string src,dst="";

char tolower(char c){
  if(c>='A' && c<='Z')
    return c + ('a'-'A');
  return c; 
}

char toupper(char c){
  if(c>'a' && c<='z')
    return c + ('A'-'a');
  return c;
}

bool isnumber(char c){
  return (c>='0' && c<='9')?1:0;
}

bool ischar(char c){
  return (c>='a' && c <='z' || c >='A' && c <='Z')?1:0;	
}

int main(){
  int p=0;
  cin >> p1 >> p2 >> p3;
  cin >> src;
  
  dst=src[0];
  for(int i=1;i<src.size()-1;i++){
    if(src[i]=='-'){
      if(isnumber(src[i-1]) && isnumber(src[i+1]) || ischar(src[i-1])&&ischar(src[i+1])){
        if(src[i+1]>src[i-1]){
          if(p3==1){ //顺序 
            for(char j=src[i-1]+1;j<src[i+1];j++){
              for(int k=1;k<=p2;k++){
                switch(p1){
                  case 1:
                    if(isnumber(j))
                      dst = dst + j;
                    else
                      dst = dst + tolower(j);
                    break;
                  case 2:
                    if(isnumber(j))
                      dst = dst + j;
                    else
                      dst = dst + toupper(j);
                    break;
                  case 3:
                    dst = dst + '*';
                    break;
                }								
              }
            }
          }else{	//逆序 
            for(char j=src[i+1]-1;j>src[i-1];j--){
              for(int k=1;k<=p2;k++){
                switch(p1){
                  case 1:
                    if(isnumber(j))
                      dst = dst + j;
                    else
                      dst = dst + tolower(j);
                    break;
                  case 2:
                    if(isnumber(j))
                      dst = dst + j;
                    else
                      dst = dst + toupper(j);
                    break;
                  case 3:
                    dst = dst + '*';
                    break;
                }								
              }
            }
          }
        }else{
          dst = dst + src[i];
        }
      }else{
        dst = dst + src[i];
      }
    }else{
      dst = dst + src[i];
    }
  }
  dst = dst + src[src.length()-1];
  
  cout << dst << endl;
  
  return 0;
}