POJ - 1845 Sumdiv

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). 

题意：求 a^b 的所有约数的和

题解：

tips：

 

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<cstring>

using namespace std;
const int mod=9901;
const int maxn=5e5+5;
typedef long long ll;
ll p[maxn];
int prime[maxn];
void is_prime() {
    int cnt=0,i,j;
    memset(prime,0,sizeof(prime));
    for(i=2; i<maxn; i++) {
        if(prime[i]==0) {
            p[++cnt]=i;
            for(j=2*i; j<maxn; j=j+i) {
                prime[j]=1;
            }
        }
    }
}
ll multi(ll A,ll n,ll k) {
    ll b=0;
    while(n>0) {
        if(n&1) {
            b=(b+A)%k;
        }
        n=n>>1;
        A=(A+A)%k;
    }
    return b;
}
ll getresult(ll A,ll n,ll k) {
    ll b=1;
    while(n>0) {
        if(n&1) {
            b=multi(b,A,k);
        }
        n=n>>1;
        A=multi(A,A,k);
    }
    return b;
}
void solve(ll A,ll B) {
    int i;
    ll ans=1;
    for(i=1; p[i]*p[i]<=A; i++) {
        if(A%p[i]==0) {
            int num=0;
            while(A%p[i]==0) {
                num++;
                A=A/p[i];
            }
            ll m=(p[i]-1)*9901;
            ans*=(getresult(p[i],num*B+1,m)+m-1)/(p[i]-1);
            ans%=9901;
        }
    }
    if(A>1) {
        ll m=9901*(A-1);
        ans *= (getresult(A,B+1,m)+m-1)/(A-1);
        ans%=9901;
    }
    cout<<ans<<endl;
}
int main() {
    ll A,B;
    is_prime();
    while(cin>>A>>B) {
        solve(A,B);
    }
    return 0;
}