Example 4.1

利用for循环，计算输出1+2+...+100的和。

• [+]

  #include <iostream>
  using namespace std;
  
  int main(){
  	int sum = 0;
  	
  	for(int i=1;i<=100;i++)
  		sum = sum + i;
  		
  	cout << sum << endl; 
  
  	return 0;
  }
  

Example 4.2

输出1~100之间所有偶数。

• [+]

  #include <iostream>
  using namespace std;
  
  int main(){
  	
  	for(int i=1;i<=100;i++)
  		if(i%2==0)
  			cout << i << " ";
  	cout << endl;
  	
  	return 0;
  }
  

Example 4.3

利用for循环，分别计算1~100中的奇数的和、偶数的和。

• [+]

  #include <iostream>
  using namespace std;
  
  int main(){
  	int odd=0,even=0;
  	for(int i=1;i<=100;i++)
  		if(i%2==0){
  			even += i;
  		}else{
  			odd += i;
  		}
  		
  	cout << "odd sum is " << odd << ", even sum is " << even << endl;
  		
  	return 0;
  }
  

• [+]

  #include <iostream>
  using namespace std;
  
  int main(){
  	int odd=0,even=0;
  	
  	for(int i=2,j=1;i<=100;i+=2,j+=2){
  		even += i;
  		odd +=j;
  	}
  
  	cout << "odd sum is " << odd << ", even sum is " << even << endl;
  	
  	return 0;
  }
  

Example 4.4

利用for循环计算n!的值。

• [+]

  #include <iostream>
  using namespace std;
  
  int main(){
  	int n;
  	long long p = 1;
  	
  	cin >> n;
  	
  	for(int i=1;i<=n;i++)
  		p = p * i;
  		
  	printf("%lld\n",p);
  	
  	return 0;
  }
  

Exercise 1

1. 求平均年龄http://noi.openjudge.cn/ch0105/01/[+]

   #include <iostream>
   using namespace std;
   
   int main(){
   	
   	int n,a;
   	double sum = 0;
   	cin >> n;
   	for (int i=0;i<n;i++){
   		cin >> a;
   		sum = sum + a;	
   	}
   	
   	printf("%.2f",sum/n);	
   }
   

2. 财务管理http://noi.openjudge.cn/ch0105/02/[+]

   #include <iostream>
   using namespace std;
   
   int main(){
   	
   	double a;
   	double sum = 0;
   	for (int i=0;i<12;i++){
   		cin >> a;
   		sum = sum + a;	
   	}
   	
   	printf("$%.2f",sum/12);	
   }
   

3. 均值http://noi.openjudge.cn/ch0105/03/[+]

   #include <iostream>
   using namespace std;
   
   int main(){
   	
   	int n;
   	double a;
   	double sum = 0;
   	cin >> n;
   	for (int i=0;i<n;i++){
   		cin >> a;
   		sum = sum + a;	
   	}
   	
   	printf("%.4f",sum/n);	
   }
   

4. 求整数的和与均值http://noi.openjudge.cn/ch0105/04/[+]

   #include <iostream>
   using namespace std;
   
   int main(){
   	
   	int n,a,sum = 0;
   	cin >> n;
   	for (int i=0;i<n;i++){
   		cin >> a;
   		sum = sum + a;	
   	}
   	printf("%d %.5f",sum,(double)sum/n);	
   	return 0;
   }
   

5. 最高的分数http://noi.openjudge.cn/ch0105/05/[+]

   #include <iostream>
   using namespace std;
   //05:最高的分数
   int main(){
   	
   	int n,a,max = 0;
   	cin >> n;
   	for (int i=0;i<n;i++){
   		cin >> a;
   		if (a>max){
   			max = a;
   		}	
   	}
   	printf("%d",max);	
   	return 0;
   }
   

6. 整数序列的元素最大跨度值http://noi.openjudge.cn/ch0105/06/[+]

   #include <iostream>
   using namespace std;
   //06:整数序列的元素最大跨度值
   int main(){
   	
   	int n,a,max = 0,min=10000;
   	cin >> n;
   	for (int i=0;i<n;i++){
   		cin >> a;
   		if (a>max){
   			max = a;
   		}
   		if(a<min){
   			min = a;
   		}
   	}
   	printf("%d",max-min);	
   	return 0;
   }
   

7. 奥运奖牌计数http://noi.openjudge.cn/ch0105/07/[+]

   #include <iostream>
   using namespace std;
   //07:奥运奖牌计数
   int main(){
   	
   	int n,g,s,c,sg=0,ss=0,sc=0;
   	cin >> n;
   	for (int i=0;i<n;i++){
   		cin >> g >> s >> c;
   		sg = sg + g;
   		ss = ss + s;
   		sc = sc + c; 
   	}
   	printf("%d %d %d %d",sg,ss,sc,sg+ss+sc);	
   	return 0;
   }
   

8. 多边形内角和http://noi.openjudge.cn/ch0105/08/[+]

   #include <iostream>
   using namespace std;
   //08:多边形内角和
   int main(){
   	
   	int n,a,sum=0;
   	cin >> n;
   	for (int i=1;i<n;i++){
   		cin >> a;
   		sum = sum + a;
   	}
   	printf("%d",(n-2)*180-sum);	
   	return 0;
   }
   

9. 奇数求和http://noi.openjudge.cn/ch0105/09/[+]

   #include <iostream>
   using namespace std;
   
   int main(){
   	int m,n;
   	int sum = 0;
   
   	cin >> m >> n;
   	for(int i=m;i<=n;i++){
   		if(i%2==1){
   			sum = sum + i;
   		}
   	}	
   	cout << sum << endl;
   	
   	return 0;
   }
   

10. 满足条件的数累加http://noi.openjudge.cn/ch0105/10/[+]

    #include <iostream>
    using namespace std;
    //10:满足条件的数累加
    int main(){
    	
    	int m,n,sum=0;
    	cin >> m >> n;
    	for (int i=m;i<=n;i++){
    		if(i%17==0){
    			sum = sum + i;
    		}
    	}
    	printf("%d",sum);	
    	return 0;
    }
    

11. 整数的个数http://noi.openjudge.cn/ch0105/11/[+]

    #include <iostream>
    using namespace std;
    //11:整数的个数
    int main(){
    	
    	int n,a,s1=0,s5=0,s10=0;
    	cin >> n;
    	for (int i=0;i<n;i++){
    		cin >> a;
    		if(a==1){
    			s1 = s1 + 1;
    		}
    		if(a==5){
    			s5 = s5 + 1;
    		}
    		if(a==10){
    			s10 = s10 + 1;
    		}
    	}
    	cout << s1 << endl;	
    	cout << s5 << endl;	
    	cout << s10 << endl;	
    	return 0;
    }
    

12. 与指定数字相同的数的个数http://noi.openjudge.cn/ch0105/12/[+]

    #include <iostream>
    using namespace std;
    
    int main()
    {
    	int N,m,a,sum=0;
    
    	cin >> N >> m;
    	for (int i=0;i<N;i++){
    		cin >> a;
    		if(a==m){
    			sum = sum + 1;
    		}
    	}
    	cout << sum << endl;	
    
    	return 0;
    }
    

13. 乘方计算http://noi.openjudge.cn/ch0105/13/[+]

    #include <iostream>
    using namespace std;
    
    int main()
    {
    	int a,n,mul=1;
    
    	cin >> a >> n;
    	for(int i=0;i<n;i++)
    	{
    		mul = mul * a;
    	}
    	cout << mul << endl;
    			
    	return 0;
    }
    

14. 人口增长问题http://noi.openjudge.cn/ch0105/14/[+]

    #include <iostream>
    #include <cmath>
    using namespace std;
    
    int main()
    {
    	double x,n;
    
    	cin >> x >> n;
    	printf("%.4f",x*pow((1+0.001),n));
    			
    	return 0;
    }
    

15. 银行利息http://noi.openjudge.cn/ch0105/15/[+]

    #include <iostream>
    using namespace std;
    
    int main()
    {
    	double R,M,Y;
    	
    	cin >> R >> M >> Y;
    	for (int i=0;i<Y;i++)
    	{
    		M = M * (1 + R/100);
    	}
    	printf("%d",(int)M);
    	
    	return 0;
    } 
    

16. 买房子http://noi.openjudge.cn/ch0105/16/[+]

    #include <iostream>
    using namespace std;
    
    int main(){
    	double sumn,hp=200,n,k;
    	int m=1;
    	cin >> n >> k;
    	sumn = n;
    	while(sumn<hp && m <40){
    		sumn = sumn + n;
    		hp = hp * (1+k/100);
    		m = m + 1;
    		
    	}
    	
    	if(m<=20)
    		cout << m << endl;
    	else
    		cout << "Impossible" << endl;
    	
    	return 0;
    }
    

17. 菲波那契数列http://noi.openjudge.cn/ch0105/17/[+]

    #include <iostream>
    using namespace std;
    
    int main()
    {
    	int k,first=1,second=1,fb;
    	
    	cin >> k;
    	if (k > 2)
    		for (int i=3;i<=k;i++)
    		{
    			fb = first + second;
    			first = second;
    			second = fb;
    		}
    	else
    		fb = 1;
    	cout << fb << endl;
    			
    	return 0;
    } 
    

18. 鸡尾酒疗法http://noi.openjudge.cn/ch0105/18/[+]

    #include <iostream>
    using namespace std;
    
    int main()
    {
    	int n;
    	double a[20][2];
    	
    	cin >> n;
    	for (int i=0;i<n;i++)
    	{
    		cin >> a[i][1] >> a[i][2];
    	}
    	
    	double x = a[0][2] / a[0][1] * 100;
    	for(int i=1;i<n;i++){
    		double y = a[i][2] / a[i][1] * 100;
    		if (y-x>5) {
    			cout << "better" << endl;	
    		}else if (x-y>5){
    			cout << "worse" << endl;	
    		}else{
    			cout << "same" << endl;	
    		}
    	}
    			
    	return 0;
    } 
    

19. 救援http://noi.openjudge.cn/ch0105/19/[+]

    #include <iostream>
    #include <cmath>
    using namespace std;
    
    int main()
    {
    	int n;
    	double a,b,c,t=0;
    	
    	cin >> n;
    	for(int i=0;i<n;i++)
    	{
    		cin >> a >> b >> c;
    		t = t + sqrt(a*a+b*b) * 2 / 50 + 1.5 * c;
    	}
    	printf("%.0f",ceil(t));
    	
    	return 0;	
    } 
    

20. 津津的储蓄计划http://noi.openjudge.cn/ch0105/22/[+]

    #include <iostream>
    using namespace std;
    
    int main()
    {
    	long p,a,r =0,sum=0;
    
    	for(int i=0;i<12;i++)
    	{
    		r = r + 300;
    		cin >> p;
    		r = r - p;
    		a = r/100;
    		sum = sum + a * 100;
    		r = r - a * 100;
    		if (r<0){
    			cout << "-" << i+1 << endl;
    			break;
    		}	 
    	}
    	if (r>0)
    		cout << sum * 1.2 + r << endl;
    	
    	return 0;	
    } 
    

21. 药房管理http://noi.openjudge.cn/ch0105/23/[+]

    #include <iostream>
    using namespace std;
    
    int main()
    {
    	long m,n,a,sum=0;
    	
    	cin >> m >>n;
    	for(int i=0;i<n;i++)
    	{
    		cin >> a;
    		if (m -a >= 0)
    			m = m - a;
    		else
    			sum = sum + 1;
    	}
    	cout << sum  << endl;
    	
    	return 0;	
    } 
    

22. 正常血压http://noi.openjudge.cn/ch0105/24/[+]

    #include <iostream>
    using namespace std;
    
    int main()
    {
    	long n,a,b,sum=0,max=0;
    	
    	cin >> n;
    	for(int i=0;i<n;i++)
    	{
    		cin >> a >> b;
    		if (a>=90 && a<=140 && b >=60 && b <=90)
    		{
    			sum = sum + 1;		
    		}else{
    			if (sum > max)
    				max = sum;
    			sum = 0;
    		}
    	}
    	if (sum > max) max = sum;
    	cout << max  << endl;
    	
    	return 0;	
    } 
    

23. 求特殊自然数http://noi.openjudge.cn/ch0105/25/[+]

    #include <iostream>
    using namespace std;
    
    int main()
    {
    	int i,j,k;
    	
    	bool flag = false;
    	for(i=0;i<7;i++)
    	{
    		for(j=0;j<7;j++)
    		{
    			for(k=0;k<7;k++){
    				if (i*9*9+j*9+k==i+j*7+k*7*7 && (i!=0 || j!=0 || k!=0))
    				{
    					cout << i*9*9+j*9+k  << endl;
    					cout << k << j << i << endl;
    					cout << i << j << k << endl; 
    				}
    			}
    		}
    	}
    	
    	return 0;	
    } 
    

24. 统计满足条件的4位数个数http://noi.openjudge.cn/ch0105/26/[+]

    #include <iostream>
    using namespace std;
    
    int main()
    {
    	int n,a,sum=0;
    	int first,second,third,forth;
    	
    	cin >> n;
    	for(int i=0;i<n;i++)
    	{
    		cin >> a;
    		forth = a % 10;
    		third = (a - forth)/10 % 10;
    		second = (a-forth - third * 10)/100 % 10;
    		first = (a - forth - third*10 - second * 100)/1000;
    		if (forth - first - second - third > 0)
    		{
    			sum = sum + 1;
    		}
    	}
    	cout << sum << endl;
    	
    	return 0;
    } 
    

25. 开关灯http://noi.openjudge.cn/ch0105/31/[+]

    #include <iostream>
    using namespace std;
    
    int main()
    {
    	long N,M;
    	bool light = true;
    
    	cin >> N >> M;
    	cout << 1 ;
    	for(int i=2;i<=N;i++)
    	{
    		light = true;	
    		for(int j=1;j<=M;j++)
    		{
    			if(i%j==0)
    			{
    				light = ! light;
    			}
    		}
    		if(!light)
    			cout << "," << i;
    		
    	}
    	
    	return 0;
    } 
    

26. 求分数序列和http://noi.openjudge.cn/ch0105/32/[+]

    #include <iostream>
    using namespace std;
    
    int main()
    {
    	int n;
    	double p=1,q=2,sum = 0,temp;
    
    	cin >> n;
    	for (int i=0;i<n;i++)
    	{
    		sum = sum + q / p;
    		temp = q;
    		q = q + p;
    		p = temp;
    	}
    	printf("%.4f",sum);
    		
    	return 0;
    } 
    

27. 计算分数加减表达式的值http://noi.openjudge.cn/ch0105/33/[+]

    #include <iostream>
    using namespace std;
    
    int main()
    {
    	int n;
    	double sum = 0;
    
    	cin >> n;
    	for (int i=1;i<=n;i++)
    	{
    		if(i%2==1)
    			sum = sum + 1.0d/i;
    		else
    			sum = sum - 1.0d/i;
    	}
    	printf("%.4f",sum);
    		
    	return 0;
    } 
    

28. 余数相同问题http://noi.openjudge.cn/math/7647/[+]

    #include <iostream>
    using namespace std;
    
    int main(){
    	int a,b,c;
    	cin >> a >> b >> c;
    	
    	for(int i=2;i<1000000;i++){
    		int t = a%i;
    		if(b%i==t && c%i==t){
    			cout << i << endl;
    			break;
    		}
    	}
    	
    	return 0;
    } 
    

29. 分苹果http://noi.openjudge.cn/math/7826/[+]

    #include <iostream>
    using namespace std;
    
    int main(){
    	int n,sum=0;
    	
    	cin >> n;
    	for(int i=1;i<=n;i++){
    		sum = sum + i;
    	}
    	cout << sum << endl;
    	
    	return 0;
    }
    

30. 求小数的某一位http://noi.openjudge.cn/math/7830/[+]

    #include <iostream>
    using namespace std;
    
    int main(){
    	int a,b,n;
    	
    	cin >> a >> b >> n;
    	
    	for(int i=1;i<=n;i++){
    		a = a%b;
    		a = a*10;
    	}
    	
    	cout << a/b << endl;
    	
    	return 0;
    }
    

31. 计算星期几http://noi.openjudge.cn/math/7831/[+]

    #include <iostream>
    #include <cstring>
    using namespace std;
    
    string wk[7]={"Sunday","Monday","Tuesday","Wednesday","Thursday","Friday","Saturday"};
    
    int main(){
    	int a,b,t=1;
    	
    	cin >> a >> b;
    	
    	for(int i=1;i<=b;i++){
    		t = t * a ;
    		t = t % 7;
    	} 
    	
    	cout << wk[t] << endl;
    	
    	return 0;
    }
    

32. 幂的末尾http://noi.openjudge.cn/math/7833/[+]

    #include <iostream>
    using namespace std;
    
    int main(){
    	int a,b,t=1;
    	
    	cin >> a >> b;
    	
    	for(int i=1;i<=b;i++){
    		t = t * a;
    		t = t % 1000;
    	}
    	
    	if(t<10)
    		cout << "00" << t << endl;
    	else if(t<100)
    		cout << "0" << t << endl;
    	else
    		cout << t << endl;
    	
    	return 0;
    }
    

Example 4.5

求s=1+2+3+...+n，当加到第几项时，s的值会超过1000？

• [+]

  #include <iostream>
  using namespace std;
  
  int main(){
  	int n=0,s=0;
  
  	while(s<=1000){
  		n++;
  		s+=n;
  	}
  	cout << n << endl;
  	
  	return 0;
  }
  

Example 4.6

求两个正整数m,n的最大公约数。

• [+]

  #include <iostream>
  using namespace std;
  
  int main(){
  	int m,n,gcd;
  	cin >> n >> m;
  	
  	gcd = m>n?n:m;
  	while(gcd>1 && (m%gcd!=0||n%gcd!=0)) gcd--;
  	cout << gcd << endl;
  
  	return 0;
  }
  

• [+]

  #include <iostream>
  using namespace std;
  
  int main(){
  	int m,n;
  	
  	cin >> n >> m;
  	int r=n % m;
  	while(r!=0){
  		n = m;
  		m = r;
  		r = n % m;
  	}
  	cout << m << endl;
  	
  	return 0;
  }
  

Example 4.7

编一程序求满足不等式1+1/2+1/3+...+1/n>=5的最小n值。

• [+]

  #include <iostream>
  using namespace std;
  
  int main(){
  	int i=0;
  	float s = 0;
  
  	while(s<5){
  		++i;
  		s+=1.0/i;
  	}
  	cout << i << endl;
  	
  	return 0;
  }
  

• [+]

  #include <iostream>
  using namespace std;
  
  int main(){
  	int i;
  	float s=0;
  	
  	for(i=1;s<5;i++){
  		s=s+1.0/i;
  	}
  	cout << i-1 << endl;
  	
  	return 0;
  }
  

Example 4.8

输入一个整数，求出他们的最小值，最大值和平均值（保留3位小数）。输入保证这些数都是不超过1000的整数。

输入样例：2 8 3 5 1 7 3 6

输出样例：1 8 4.375

• [+]

  #include <iostream>
  using namespace std;
  
  int main(){
  	int x,n=0,min=10000,max=0,s=0;
  	
  	while(scanf("%d",&x)==1){
  		s+=x;
  		min = x<min?x:min;
  		max = x>max?x:max;
  		n++;
  	}
  
  	printf("%d %d %.3lf",min,max,(double)s/n);
  	
  	return 0;
  }
  

• [+]

  #include <iostream>
  #define INF 100000000
  using namespace std;
  
  int main(){
  	int x,n=0,min=INF,max=-INF,s=0;
  	while(scanf("%d",&x)!=EOF){
  		s+=x;
  		min = x<min?x:min;
  		max = x>max?x:max;
  		n++;
  	}
  
  	printf("%d %d %.3lf",min,max,(double)s/n);
  
  	return 0;
  }
  

Example 4.9

对于求两个正整数m，n的最大公约数可以用do-while实现。

• [+]

  #include <iostream>
  using namespace std;
  
  int main(){
  	int m,n,r;
  
  	cin >> m >> n;
  	do{
  		r = m % n;
  		m = n;
  		n = r;
  	}while(r!=0);
  	cout << m << endl;
  	
  	return 0;
  }
  

Example 4.10

求1992个1992的乘积的末两位数是多少？

• [+]

  #include <iostream>
  using namespace std;
  
  int main(){
  	int a=1,t=0;
  	
  	do{
  		++t;
  		a=(a*1992)%100;
  	}while(t!=1992);
  	cout << a << endl;
  	
  	return 0;
  }
  

Example 4.11

校体操队到操场集合，排成每行2人，最后多出1人；排成每行3人，也多出1人，分别按每行排4,5,6人，都多出1人；当排成每行7人时，正好不多。求校体操队至少多少人？

• [+]

  #include <iostream>
  using namespace std;
  
  int main(){
  	int x = 0;
  	bool yes;
  	
  	do{
  		yes = true;
  		x+=7;
  		if(x%2!=1) yes = false;
  		if(x%3!=1) yes = false;
  		if(x%4!=1) yes = false;
  		if(x%5!=1) yes = false;
  		if(x%6!=1) yes = false;
  	}while(yes==false);
  	
  	cout << x << endl;
  	
  	return 0;
  }
  

Exercise 2

1. 球弹跳高度的计算http://noi.openjudge.cn/ch0105/20/[+]

   #include <iostream>
   using namespace std;
   
   int main()
   {
   	double h,sum=0;
   	
   	cin >> h;
   	
   	sum = h;
   	for(int i=1;i<10;i++)
   	{
   		sum = sum + h;
   		h = h / 2;
   	}
   	cout << sum << endl;
   	cout << h/2 << endl;
   	
   	return 0;	
   } 
   

2. 角谷猜想http://noi.openjudge.cn/ch0105/21/[+]

   #include <iostream>
   using namespace std;
   
   int main()
   {
   	long long int N;
   	cin >> N;
   	
   	while(N>1)
   	{
   		if(N%2==1)
   		{
   			cout << N << "*3+1=" << N * 3 + 1 << endl;
   			N = N * 3 + 1;
   		}else{
   			cout << N << "/2=" << N / 2 << endl;
   			N = N / 2; 
   		} 
   	}
   	cout << "End";
   	
   	return 0;	
   } 
   

3. 级数求和http://noi.openjudge.cn/ch0105/27/[+]

   #include <iostream>
   using namespace std;
   
   int main()
   {
   	long K,n=0;
   	double sn = 0;
   	
   	cin >> K;
   	while(sn<=K)
   	{
   		n = n + 1;
   		sn = sn + 1.0d/n;
   	}
   	cout << n << endl;
   			
   	return 0;
   } 
   

4. 分离整数的各个数位http://noi.openjudge.cn/ch0105/28/[+]

   #include <iostream>
   using namespace std;
   
   int main()
   {
   	long long int n,r;
   	
   	cin >> n;
   	while(n>0)
   	{
   		r = n % 10;
   		cout << r;
   		n = (n - r)/10;
   		if (n>0)
   		{
   			cout << " ";
   		} 
   	}
   	cout << endl;
   				
   	return 0;
   } 
   

5. 数字反转http://noi.openjudge.cn/ch0105/29/[+]

   #include <iostream>
   using namespace std;
   
   int main()
   {
   	long long int n,i=0;
   	
   	cin >> n;
   	while(n!=0)
   	{
   		i = i*10 + n % 10;
   		n = n / 10;
   	}
   	printf("%d",i);
   	
   	return 0;
   } 
   

6. 含k个3的数http://noi.openjudge.cn/ch0105/30/[+]

   #include <iostream>
   using namespace std;
   
   int main()
   {
   	long m,k,sum=0;
   	
   	cin >> m >> k;
   	if (m % 19 ==0)
   	{
   		while(m!=0)
   		{
   			if (m % 10==3)
   			{
   				sum = sum + 1;
   			}
   			m = m / 10;
   		}
   		if (sum==k)
   			cout << "YES" << endl;
   		else
   			cout << "NO" << endl;
   		
   	}
   	else
   		cout << "NO" << endl;
   	
   	return 0;
   } 
   

Example 4.12

求s=1!+2!+3!+...+10!

• [+]

  #include <iostream>
  using namespace std;
  
  int main(){
  	int s=0;
  	
  	for(int i=1;i<=10;i++){
  		int t=1;
  		for(int j=1;j<=i;j++)
  			t=t*j;
  		s+=t;
  	}
  	cout << s << endl;
  
  	return 0;
  }
  

• [+]

  #include <iostream>
  using namespace std;
  
  int main(){
  	int t=1,s=0;
  	
  	for(int i=1;i<=10;i++){
  		t*=i;
  		s+=t;
  	}
  	cout << s << endl;
  	
  	return 0;
  }
  

Example 4.13

对于给定的自然数n(n<20)，在屏幕上输出仅由"*"构成的n行的直角三角形。

• [+]

  #include <iostream>
  using namespace std;
  
  int main(){
  	int n;
  	
  	cin >> n;
  	for(int i=1;i<=n;i++){
  		for(int j=1;j<=i;j++){
  			cout << "*";
  		}
  		cout << endl; 
  	}
  
  	return 0;
  }
  

Example 4.14

百钱买百鸡问题。鸡翁一，值钱五，鸡母一，值钱三，鸡雏三，值钱一，百钱买百鸡，问鸡翁、鸡母、鸡雏各几何？

• [+]

  #include <iostream>
  using namespace std;
  
  int main(){
  	int x,y,z;
  	
  	for(x=0;x<=100/5;x++){
  		for(y=0;y<=100/3;y++){
  			for(z=0;z<=3*100;z++){
  				if(5*x+3*y+z/3==100 && x+y+z==100&&z%3==0)
  					cout << x << " " << y << " " << z << endl;
  			}
  		}
  	}
  
  	return 0;
  }
  

• [+]

  #include <iostream>
  using namespace std;
  
  int main(){
  	int x,y,z;
  	for(x=0;x<=100/5;x++){
  		for(y=0;y<=100/3;y++){
  			z = 100 -x -y;
  			if(5*x+3*y+z/3==100 && z%3==0)
  				cout << x << " " << y << " " << z << endl;
  		}
  	}
  
  	return 0;
  }
  

Example 4.15

求100~999中的水仙花数。若三位数ABC，ABC=A³+B³+C³，则称ABC为水仙花数。
例如153，1³+2³+3³=1+125+27=153，则153是水仙花数。

• [+]

  #include <iostream>
  #include <iomanip>
  using namespace std;
  
  int main(){
  	
  	for(int a=1;a<=9;a++){
  		for(int b=0;b<=9;b++){
  			for(int c=0;c<=9;c++){
  				if(a*a*a+b*b*b+c*c*c==a*100+b*10+c)
  					cout << setw(6) << a*100+b*10+c;
  			}
  		}
  	}
  
  	return 0;
  }
  

• [+]

  #include <iostream>
  #include <iomanip>
  using namespace std;
  
  int main(){
  
  	for(int m=100;m<=999;m++){
  		int a=m/100;
  		int b=(m%100)/10;
  		int c = m%10;
  		if(a*a*a+b*b*b+c*c*c==m)
  			cout << setw(6) << m;
  	}
  	cout << endl;
  	
  	return 0;
  }
  

Example 4.16

输出100~200中所有的素数

• [+]

  #include <iostream>
  #include <cmath>
  using namespace std;
  
  int main(){
  	
  	for(int i=100;i<=200;i++){
  		int x=2;
  		while(x<=floor(sqrt(i))&&(i%x!=0))
  			x++;
  		if(x>floor(sqrt(i)))
  			cout << i << "\t";
  	}
  	cout << endl;
  
  	return 0;
  }
  

Example 4.17

输出所有形如aabb的四位完全平方数（即前两位数字相等，后两位数字也相等）。

• [+]

  #include <iostream>
  #include <cmath>
  using namespace std;
  
  int main(){
  	
  	for(int i=1;i<=9;i++){
  		for(int j=0;j<=9;j++){
  			int n=i*1000+i*100+j*10+j;
  			int m = sqrt(n);
  			if(n==m*m)
  				cout << n << endl;		
  		}
  	}
  
  	return 0;
  }
  

• [+]

  #include <iostream>
  using namespace std;
  
  int main(){
  	
  	for(int x=32;;x++){
  		int n=x*x;
  		if(n<1000) continue;
  		if(n>9999) break;
  		int hi=n/100,lo=n%100;
  		if(hi/10==hi%10 && lo/10==lo%10)
  			cout << n << endl;
  	}
  
  	return 0;
  }
  

Example 4.18

把一个合数分解成若干个质因数乘积的形式（即求质因数的过程）叫做分解质因数。
分解质因数（也称分解素因素）只针对合数。

• [+]

  #include <iostream>
  using namespace std;
  
  int main(){
  	int n,i=2;
  	
  	cin >> n;
  	cout << n << "=";
  	do{
  		while(n%i==0){
  			cout << i;
  			n/=i;
  			if(n!=1) 
  				cout << "*";
  		}
  		i++;
  	}while(n!=1);
  	cout <<endl;
  	
  	return 0;
  }
  

Example 4.19

输入n，计算s=1!+2!+3!+...+n!的末6位（不含前导0）。n<=10⁶，n!表示前n个正整数之积。

• [+]

  #include <iostream>
  using namespace std;
  
  int main(){
  	int n,s=0;
  	
  	cin >> n;
  	for(int i=1;i<=n;i++){
  		int factorial=1;
  		for(int j=1;j<=i;j++){
  			factorial*=j;
  		}
  		s += factorial;
  	}
  	cout << s%1000000 << endl;
  	
  	return 0;
  }
  

• [+]

  #include <iostream>
  #include <ctime>
  using namespace std;
  
  int main(){
  	const int MOD = 1000000;
  	int n,s=0;
  
  	cin >> n;
  	for(int i=1;i<=n;i++){
  		int factorial = 1;
  		for(int j=1;j<=i;j++){
  			factorial = (factorial * j )% MOD; 
  			s = (s+factorial) % MOD; 
  		}
  	}
  	cout << s << endl;
  	printf("Time used = %.2lf\n",(double)clock()/CLOCKS_PER_SEC);
  	
  	return 0;
  }
  

Exercise 4

1. 求阶乘的和http://noi.openjudge.cn/ch0105/34/[+]

   #include <iostream>
   using namespace std;
   
   int main()
   {
   	int n;
   	long long int sum = 0,product=1;
   
   	cin >> n;
   	for (int i=1;i<=n;i++)
   	{
   		product = 1;
   		for(int j=1;j<=i;j++)
   		{
   			product = product * j;
   		}
   		sum = sum + product;
   	}
   	cout << sum << endl; 
   		
   	return 0;
   } 
   

2. 求出e的值http://noi.openjudge.cn/ch0105/35/[+]

   #include <iostream>
   using namespace std;
   
   int main(){
   	
   	int n;
   	double e = 1,p;
   	cin >> n;
   	for (int i=1;i<=n;i++){
   		p = 1;
   		for(int j=1;j<=i;j++)
   		{
   			p = p * j;
   		}
   		e = e + 1/p;
   	}
   	
   	printf("%.10f",e);	
   }
   

3. 计算多项式的值http://noi.openjudge.cn/ch0105/36/[+]

   #include <iostream>
   #include <cmath>
   using namespace std;
   
   int main(){
   	
   	long int n;
   	double x,f = 1,p;
   	cin >> x >> n ;
   	f = f + x;
   	for (int i=2;i<=n;i++){
   		f = f + pow(x,i);
   	}
   	
   	printf("%.2f",f);	
   }
   

4. 雇佣兵http://noi.openjudge.cn/ch0105/37/[+]

   #include <iostream>
   using namespace std;
   
   int main(){
   	
   	int M,N,X;
   	
   	cin >> M >> N >> X ;
   	
   	while(N <=M && X * N >=M)
   	{
   		if (M%N==0)
   		{
   			X = X - M / N;
   		}
   		else
   		{
   			X = X - M / N -1;
   		}
   		N = N + M/N;
   
   	}
   	printf("%d",N);
   	
   	return 0;	
   }
   

5. 计算多项式的导函数http://noi.openjudge.cn/ch0105/38/[+]

   #include <iostream>
   using namespace std;
   
   int main(){
   	
   	int n,c[100];
   	
   	cin >> n ;
   
   	for(int i=n;i>=0;i--){
   		cin >> c[i];
   		c[i] = c[i] * i;
   	}
   	
   	cout << c[n];
   	for(int i=n-1;i>0;i--)
   	{
   		cout << " " << c[i];
   	}
   	cout << endl;
   	
   	return 0;	
   }
   

6. 与7无关的数http://noi.openjudge.cn/ch0105/39/[+]

   #include <iostream>
   using namespace std;
   
   int main(){
   	
   	int n,sum=0,a;
   	
   	cin >> n ;
   	for(int i=1;i<=n;i++){
   		if(i%7!=0){
   			a = i;
   			while(a>0)
   			{
   				if (a % 10 == 7)
   				{
   					break;					
   				}
   				a = a / 10;	
   			}
   			if (a ==0){
   				sum = sum + i * i;
   			}
   		}
   	}
   	cout << sum << endl;
   	
   	return 0;	
   }
   

7. 数1的个数http://noi.openjudge.cn/ch0105/40/[+]

   #include <iostream>
   using namespace std;
   
   int main(){
   	
   	int n,sum=0,a;
   	
   	cin >> n ;
   	for(int i=1;i<=n;i++){
   		a = i;
   		while(a>0)
   		{
   			if (a % 10 == 1)
   			{
   				sum = sum + 1;
   			}
   			a = a / 10;	
   		}
   	}
   	cout << sum << endl;
   	
   	return 0;	
   }
   

8. 数字统计http://noi.openjudge.cn/ch0105/41/[+]

   #include <iostream>
   using namespace std;
   
   int main(){
   	
   	int L,R,sum=0,a;
   	
   	cin >> L >> R ;
   	for(int i=L;i<=R;i++){
   		a = i;
   		while(a>0)
   		{
   			if (a % 10 == 2)
   			{
   				sum = sum + 1;
   			}
   			a = a / 10;	
   		}
   	}
   	cout << sum << endl;
   	
   	return 0;	
   }
   

9. 画矩形http://noi.openjudge.cn/ch0105/42/[+]

   #include <iostream>
   using namespace std;
   
   int main(){
   	
   	int h,w,fill;
   	char c;
   	
   	cin >> h >> w >> c >> fill ;
   	
   	for (int i=0;i<w;i++) cout << c;
   	cout << endl;
   	
   	for (int i=0;i<h-2;i++){
   		cout << c;
   		for(int j=0;j<w-2;j++)
   			if (fill==0) 
   				cout << " ";
   			else
   				cout << c;
   		cout << c << endl;
   	}
   	for (int i=0;i<w;i++) cout << c;
   	cout << endl;
   	
   	
   	return 0;	
   }
   

10. 质因数分解http://noi.openjudge.cn/ch0105/43/[+]

    #include <iostream>
    using namespace std;
    
    int main(){
    	
    	long long int n,j,t;
    	
    	cin >> n ;
    	for (int i=2;i*i<n;i++){
    		if(n % i ==0 ){
    			cout << n / i << endl;
    			break;
    		}
    	}
    	
    	
    	return 0;	
    }
    

11. 第n小的质数http://noi.openjudge.cn/ch0105/44/[+]

    #include <iostream>
    #include <cmath>
    using namespace std;
    
    int main(){
    	long n,c=2,i=3,j;
    	bool flag = false;
    	cin >> n ;
    	if (n==1)
    		i = 2;
    	else
    	while(c<n)
    	{
    		i = i + 2;
    		flag = true;
    		for(j=2;j<=sqrt(i);j++)
    			if (i % j==0)
    			{
    				flag = false;
    				break;
    			}
    		if(flag)
    		{
    			c = c + 1;
    		}
    	}
    	cout << i << endl;
    	
    	return 0;	
    }
    

12. 金币http://noi.openjudge.cn/ch0105/45/[+]

    #include <iostream>
    using namespace std;
    
    int main(){
    	long n,coin=1,sum=0;
    
    	cin >> n ;
    	
    	int count = 1;
    	while(count <= n)
    	{
    		for(int j=1;j<=coin&&count <=n;j++){
    			sum = sum + coin;
    			count = count + 1;
    		}
    		coin = coin + 1;
    	}
    	cout << sum << endl;
    	
    	return 0;	
    }
    

13. 不定方程求解http://noi.openjudge.cn/math/7650/[+]

    #include <iostream>
    using namespace std;
    
    int main(){
    	int a ,b ,c ,ans=0;
    	cin >> a >> b >> c;
    	
    	for(int i=0;i<=c/a;i++)
    		for(int j=0;j<=c/b;j++)
    			if(a*i+b*j==c)
    				ans = ans + 1;
    	
    	cout << ans << endl;
    	
    	return 0;
    }
    

参考文献

1. 董永建，信息学奥数一本通（C++）第五版。
2. http://noi.openjudge.cn

辽师张大为@https://daweizh.github.io/csp/