poj 1845 Sumdiv (算数基本定理+逆元)

输入a和b,求a^b的所有因子之和。

#include <iostream>
#define ll long long

using namespace std;

const int mod=9901;

ll a,b;

int quick(ll a,ll b,ll m)
{
    ll res,t;
    res=1;
    t=a%m;
    while(b){
        if(b&1){
            res=res*t%m;
        }
        t=t*t%m;
        b>>=1;
    }
    return res;
}

int main()
{
    ios::sync_with_stdio(false);cin.tie(0);
    cin>>a>>b;
    if(b==0||a<=1) {cout<<"1"<<endl;return 0;}
    ll ans=1;
    int s;
    for(int i=2;i*i<=a;i++){
        s=0;
        while(a%i==0){
            a/=i;
            s++;
        }
        if((i-1)%mod==0)ans=ans*(quick(i,s*b+1,mod*(i-1))-1)/(i-1)%mod;
        else  ans=ans*(quick(i,s*b+1,mod)-1)*quick(i-1,mod-2,mod)%mod;
    }
    if(a>1){
        if((a-1)%mod==0)ans=ans*(quick(a,b+1,mod*(a-1))-1)/(a-1)%mod;
        else  ans=ans*(quick(a,b+1,mod)-1)*quick(a-1,mod-2,mod)%mod;
    }
    cout<<(ans+mod)%mod<<endl;
    return 0;
}
原文地址:https://www.cnblogs.com/Fy1999/p/9700917.html