poj1845

题意：求a^b的所有约数的和对9901取余。

分析：我们转化为a^b的所有质因子的等比数列的成积，例如100^1，转化为(1+2+4) * (1 + 5 ＋ 25)。由于a^b的质因子与a的质因子相同，只是每个的数量是a的质因子的b倍。具体做法是先求所有素数，求a的所有质因子，对于每个质因子求num[i]*b＋1项的等比数列。并求乘积。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
usingnamespace std;

#define maxn 10000
#define w 9901

int a, b;

boolis[maxn];
int prm[maxn];
int fac[maxn], num[maxn];

int getprm(int n)
{
    int i, j, k =0;
    int s, e = (int) (sqrt(0.0+ n) +1);
    memset(is, 1, sizeof(is));
    prm[k++] =2;
    is[0] =is[1] =0;
    for (i =4; i < n; i +=2)
        is[i] =0;
    for (i =3; i < e; i +=2)
        if (is[i])
        {
            prm[k++] = i;
            for (s = i *2, j = i * i; j < n; j += s)
                is[j] =0;
        }
    for (; i < n; i +=2)
        if (is[i])
            prm[k++] = i;
    return k;
}

int power(int a, int n)
{
    int ret =1;
    int m = a % w;
    while (n)
    {
        if (1& n)
            ret *= m;
        ret %= w;
        n >>=1;
        m *= m;
        m %= w;
    }
    return ret;
}

int cal(int d, int num)
{
    d %= w;
    if (num ==0)
        return1;
    if (num ==1)
    {
        return (d +1) % w;
    }
    int mid;
    if (num &1)
    {
        mid = power(d, num /2+1);
        return (mid +1) * cal(d, num /2) % w;
    }
    int x = power(d, num /2);
    mid = power(d, num /2+1);
    return ((mid +1) * cal(d, num /2-1) + x) % w;
}

int main()
{
    //freopen("t.txt", "r", stdin);
    scanf("%d%d", &a, &b);
    int n = getprm(int(sqrt(a)) +1);
    int temp =0;
    for (int i =0; i < n; i++)
        if (a % prm[i] ==0)
        {
            fac[temp] = prm[i];
            while (a % prm[i] ==0)
            {
                a /= prm[i];
                num[temp]++;
            }
            temp++;
        }
    n = temp;
    int ans =1;
    for (int i =0; i < n; i++)
        ans = ans * cal(fac[i], num[i] * b) % w;
    if (a !=1)
        ans = ans * cal(a, b) % w;
    printf("%d\n", ans);
    return0;
}