poj 1845(等比数列前n项和及快速幂)

最新推荐文章于 2021-07-08 09:30:58 发布

原创最新推荐文章于 2021-07-08 09:30:58 发布 · 913 阅读

0 ·

CC 4.0 BY-SA版权

数学专栏收录该内容

19 篇文章

订阅专栏

本文针对Sumdiv问题进行了详细的解析，该问题是求两个自然数A和B的所有自然数因子之和S，并计算S对9901取模的结果。输入包含两个自然数A和B，输出为S对9901取模后的结果。文章提供了AC代码，采用快速幂运算及等比数列求和的方法来高效解决此问题。

Sumdiv

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 13959		Accepted: 3433

Description

Consider two natural numbers A and B. Let S be the sum of all natural divisors of A^B. Determine S modulo 9901 (the rest of the division of S by 9901).

Input

The only line contains the two natural numbers A and B, (0 <= A,B <= 50000000)separated by blanks.

Output

The only line of the output will contain S modulo 9901.

Sample Input

2 3

Sample Output

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).

Source

Romania OI 2002

思路看：

http://hi.baidu.com/necsinmyway/item/9f10b6d96c5068fbb2f77740

AC代码：

#include<iostream>
using namespace std;
#define LL long long
LL pow_mod(LL a,LL n,int mod){         //快速幂
    LL r=1;
    LL base=a;
    while(n){
        if(n&1)
            r=r*base%mod;
        base=base*base%mod;
        n>>=1;
    }
    return r%9901;
}
LL sum(LL a,LL b,LL mod){             //二分求等比数列前N项和
    if(b==0)
        return 1;
    if(b%2==1)
        return (sum(a,b/2,mod)*(pow_mod(a,b/2+1,mod)+1))%mod;
    else
        return (sum(a,b-1,mod)+pow_mod(a,b,mod))%mod;
}
int main(){
    LL a,b;
    LL ans;
    while(cin>>a>>b){
        ans=1;
        for(LL i=2;i*i<=a;i++){           //将a分解为质数的乘积
            if(a%i==0){
                LL s=0;
                while(a%i==0){
                    s++;
                    a/=i;
                }
                ans=ans*sum(i%9901,b*s,9901)%9901;
            }
        }
        if(a>=2){
            ans=ans*sum(a%9901,b,9901)%9901;
        }
        cout<<ans<<endl;
    }
    return 0;
}