poj 1845(等比數列前n項和及快速冪)

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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

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). 

Source

Romania OI 2002

思路看：

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

AC代碼：

#include<iostream>using namespace std;#define LL long longLL pow_mod(LL a,LL n,int mod){         //快速冪    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){             //二分求等比數列前N項和    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++){           //將a分解為質數的乘積            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;}