数学基础

1. 等差数列

1. $a_n = a_1 + (n-1) \times d$
2. $S_n = na_1 + \frac{n(n-1)d}{2}, n \in N^*$

其中，$a_1$ 为首项，$a_n$ 为末项，公差为d，前n项和为$S_n$。

2. 等比数列

1. $a_n = a_1 \cdot q^{(n-1)}$
2. $S_n = \frac{a_1(1-q^n)}{1-q}(q\neq1)$

其中，$a_1$ 为首项，$a_n$ 为末项，公比为d，前n项和为$S_n$。

3. 取模运算

3.1 四则运算（除法除外）

1. $(a + b) \% p = (a \% p + b \% p) \% p$
2. $(a – b) \% p = (a \% p – b \% p) \% p$
3. $(a * b) \% p = (a \% p * b \% p) \% p$
4. $(a^b) \% p = ((a \% p)^b) \% p$

3.2 结合律

1. $((a + b) \% p + c) \% p = (a + (b+c) \% p) \% p$
2. $((a \times b) \% p \times c) \% p = (a \times (b \times c) \% p) \% p$

3.3 交换律

1. $(a + b) \% p = (b+a) \% p$
2. $(a \times b) \% p = (b \times a) \% p$

3.4 分配律

1. $((a + b) \% p \times c) \% p = ((a \times c) \% p + (b \times c) \% p) \% p$

3.5 重要定理

1. 若$a≡b (\% p)$，则对于任意的c，都有$(a + c) ≡ (b + c) (\%p)$
2. 若$a≡b (\% p)$，则对于任意的c，都有$(a \times c) ≡ (b \times c) (\%p)$
3. 若$a≡b (\% p) ，c≡d (\% p)$，则
   1. $(a + c) ≡ (b + d) (\%p)$
   2. $(a – c) ≡ (b – d) (\%p)$
   3. $(a \times c) ≡ (b \times d) (\%p)$
   4. $(a / c) ≡ (b / d) (\%p)$

4. 幂指

$ans=base^{power}$，例如$2^{1000000}$

4.1 递推

long long power1(long long base,long long power,long long mod){
	long long result = 1;
	
	for(int i=1;i<=power;i++)
		result = ((result%mod)*(base%mod))%mod;
	
	return result;
}

4.2 快速幂

1. $3^{10}=3\times3\times3\times3\times3\times3\times3\times3\times3\times3$
2. $3^{10}=(3\times3)\times(3\times3)\times(3\times3)\times(3\times3)\times(3\times3)$
3. $3^{10}=(3\times3)^5$
4. $3^{10}=9^{5}$
5. $9^{5}=9^{4}\times9^{1}$
6. $9^{5}=81^{2}\times9^{1}$
7. $9^{5}=5661^{1}\times9^{1}$

long long power2(long long base,long long power,long long mod){
	long long result = 1;
	
	while(power>0){
		if(power%2==0){
			power = power/2;
			base = ((base%mod)*(base%mod))%mod;
		}else{
			power = power -1;
			result = ((result%mod)*(base%mod))%mod;
			power = power/2;
			base = ((base%mod)*(base%mod))%mod;
		}
	}	
	return result;
}

long long power3(long long base,long long power,long long mod){
	long long result = 1;
	
	while(power>0){
		if(power%2==1){
			result = ((result%mod)*(base%mod))%mod;
		}
		power = power/2;
		base = ((base%mod)*(base%mod))%mod;
	}
		
	return result;
}

long long power4(long long base,long long power,long long mod){
	long long result = 1;
	
	while(power>0){
		if(power&1){
			result = ((result%mod)*(base%mod))%mod;
		}
		power >>=1;
		base = ((base%mod)*(base%mod))%mod;
	}
		
	return result;
}

5. 约数个数和约数和

5.1 基本定理

任何大于1的整数A都可以分解成若干质数的乘积。如果不考虑这些质数次序，A可以写成标准分解式：
$A=P_1^{a1} \times P_2^{a2} \times P_3^{a3} \times \cdots \times P_n^{an}$，
其中 $P_1 \lt P_1 \lt P_1 \lt \ldots \lt P_n$ 为质数，$a_i$为非负整数$i = 1,2,\cdots,n$。

5.2 约数个数公式

若 $A=P_1^{a1} \times P_2^{a2} \times P_3^{a3} \times \cdots \times P_n^{an}$ 为标准公式，则A的所有约数（包括1和本身）的个数等于：
$(a_1+1)\times (a_2+1)\times (a_3+1) \times \cdots \times (a_n+n)$

5.3 约数和公式

若 $A=P_1^{a1} \times P_2^{a2} \times P_3^{a3} \times \cdots \times P_n^{an}$ 为标准公式，则A的所有约数（包括1和本身）的和等于：
$(1+p_1+p_1^2+\cdots+p_1^{a1}) \times (1+p_2+p_2^2+\cdots+p_2^{a2}) \times (1+p_3+p_3^2+\cdots+p_3^{a3}) \times \cdots \times (1+p_n+p_n^2+\cdots+p_n^{an})$

1. 递推求和O(n)[+]

   #include <iostream>
   using namespace std;
   
   int a1=1,q,n,sum=1,mod=9901,acc=1;
   
   int main(){
   
   	cin >> q >> n;
   	for(int i=1;i<=n;i++){
   		acc=((acc%mod)*q)%mod;
   		sum = (sum%mod + acc%mod)%mod;
   	}
   	cout << sum << endl;
   	
   	return 0; 
   }
   

2. 分治求和O(log²n)
   1. n为奇数时
      $(1+p_1+p_1^2+p_1^3+p_1^4+p_1^5+p_1^6+p_1^7)\%9901$
      $=(1+p_1+p_1^2+p_1^3+p_1^4(1+p_1+p_1^2+p_1^3))\%9901$
      $=\underbrace{(1+p_1+p_1^2+p_1^3)}_{继续递归} \times \underbrace{(1+p_1^4)}_{直接计算}\%9901$
   2. n为偶数时
      $(1+p_1+p_1^2+p_1^3+p_1^4+p_1^5+p_1^6)\%9901$
      $=(1+p_1+p_1^2)+p_1^3+p_1^4(1+p_1+p_1^2))\%9901$
      $=(\underbrace{(1+p_1+p_1^2)}_{继续递归} \times \underbrace{(1+p_1^4)}_{直接计算}+\underbrace{p_1^3}_{直接计算})\%9901$
   3. [+]

   #include <iostream>
   #include "power4.h"
   using namespace std;
   
   long long sum(long long a, long long n,long long mod){
   	if(n ==1) return a;
   	long long t = sum(a,n/2,mod);
   	if(n&1){
   		long long cur = power4(a,n/2+1,mod);
   		t = (t+t*cur%mod)%mod;
   		t = (t+cur)%mod;
   	}else{
   		long long cur = power4(a,n/2,mod);
   		t=(t+t*cur%mod) % mod;
   	}
   	return t;
   }
   
   int main(){
   	
   	cout << 1+sum(2,3,9901);
   	
   	return 0;
   }
   

5.4 分解质因数

• [+]

  #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;
  }
  

5.5 POJ1845 Sumdiv

Consider two natural numbers A and B. Let S be the sum of all natural divisors of $A^B$.

Input

The only line contains the two natural numbers A and B, (0 <= A,B <= 500000000) separated by blanks.

Output

The only line of the output will contains S modulo 9001.

Sample Input

2 3

Sample Output

15

HINT

$2^3 = 8$

The natural divisors of 8 are: 1,2,4,8. Their sum is 15.

15 modulo 9901 is 15 (that should be output).

Analysis

$(1+p_1+p_1^2+\cdots+p_1^{a_1B}) \times (1+p_2+p_2^2+\cdots+p_2^{a_2B}) \times (1+p_3+p_3^2+\cdots+p_3^{a_3B}) \times \cdots \times (1+p_n+p_n^2+\cdots+p_n^{a_nB})$

Code

• [+]

  #include<iostream>
  #include<cstdio>
  #define size 10000
  #define mod 9901
  using namespace std;
  
  int p[size],n[size];
  
  //快速幂 
  long long pow3(long long int a,long long int b ){
  		long long int r = 1, base = a;
  		while( b != 0 ){
  				if( b & 1 ) 
  			r =(r * base) % mod;
  				base =(base * base) % mod;
  				b >>= 1;
  		}
  		return r;
  }
  
  //二分递归求解等比数列之和
  long long sum(long long p,long long n){
  		if(n==0)
  				return 1;
  		if(n%2)
  				return ((sum(p,n/2) % mod) * (1+pow3(p,n/2+1))% mod )% mod;
  		else
  				return (((sum(p,n/2-1) % mod) * (1+pow3(p,n/2+1))% mod ) % mod + pow3(p,n/2) %mod)%mod;
  }
  
  int main(){
  		int A,B;
  		scanf("%d%d",&A,&B)； 
  				int js=0;
  				//质因数分解A，根号法和递归法
  				for(int i=2;i*i<=A;){
  						if(A%i==0){
  								p[js]=i;   //质因数 
  								n[js]=0;	
  								while(A%i==0){	//质因数次数 
  										n[js]++;
  										A/=i;
  								}
  								js++;
  						}
  						if(i==2) i++;
  						else i+=2;
  				}
  				//A本身就是素数的
  				if(A!=1){
  						p[js]=A;
  						n[js++]=1;
  				}
  
  				long long ans=1;
  				for(int i=0;i<js;i++){
  						ans=( ans * sum(p[i],n[i]*B)%mod )%mod;
  				}
  				
  				cout<<ans<<endl;
  
  		return 0;
  }