A^B的所有因子和对9901求模--POJ 1845 Sumdiv 逆元 费马小定理 Trick

资料来源：https://blog.csdn.net/danliwoo/article/details/48877783

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A^B的所有因子和对9901求模

原题见POJ 1845

原以为就是循规蹈矩的将A^B算出来，将大数分解因子，老老实实求和再大数求模，大体能明白一点为啥要这么算了，细看还是不明白。。。

代入p=9901x+1p=9901x+1，则原式即为n+1n+1特判一下这种情况即可！

附code

/*--------------------------------------------
 * File Name: POJ 2992
 * Author: Danliwoo
 * Mail: [email protected]
 * Created Time: 2015-10-03 09:53:42
--------------------------------------------*/

#include <cstdio>
#include <iostream>
#include <cstring>
#include <queue>
#include <algorithm>
#include <cmath>
using namespace std;
#define N 7100
int num[N], prim[N], fq[N], cnt = 0;

int extend_Euclid(int a, int b, int &x, int &y){
    if(b==0){
    x = 1; y = 0;
    return a;
    }
    int r = extend_Euclid(b, a%b, y, x);
    y -= a/b*x;
    return r;
}
int anti(int a, int m)
{
    int x, y;
    extend_Euclid(a, m, x, y);
    return (x%m+m)%m;
}
void set()
{
    for(int i = 0;i < 7100;i++) num[i] = 1;
    for(int i = 2;i < 7100;i++) if(num[i])
    {
        fq[i-1] = anti(i-1, 9901);
        prim[cnt++] = i;
        for(int j = i*i;j < 7100;j += i)
            num[j] = 0;
    }
}
int po(int a, int k, int m)
{
    if(k == 0) return 1;
    if(k == 1) return a;
    int t = po(a, k/2, m);
    t = t*t%m;
    if(k & 1LL)
        t *= a;
    return t%m;
}
int main()
{
    set();
    int a, b, m = 9901;
    while(~scanf("%d%d", &a, &b))
    {
        if(a == 0)
        {
            printf("0\n");
            continue;
        }
        int ans = 1;
        int t = a;
        for(int i = 0;i < cnt && prim[i]*prim[i] <= a;i++) if(t%prim[i] == 0)
        {
            int s = 0;
            while(t % prim[i] == 0)
            {
                t /= prim[i];
                s++;
            }
            ans = ans*fq[prim[i]-1]%m*(po(prim[i], b%(m-1)*s%(m-1)+1, m)-1)%m;
        }
        if(t != 1)
        {
            if(t % m == 1)
                ans = ans*(b%m+1)%m;
            else
                ans = ans*anti(t-1, m)%m*(po(t%m, b%(m-1)+1, m)-1)%m;
        }
        printf("%d\n", (ans%m+m)%m);
    }
    return 0;
}