因子和

题目描述

输入两个正整数a和b，求ab的因子和。结果太大，只要输出它对9901的余数。

输入输出格式

输入格式：

仅一行，为两个正整数a和b(0≤a,b≤50000000)。

输出格式：

a^b的因子和对9901的余数。

输入输出样例

输入样例#1

2 3

输出样例#1

15

题解

根据约数和定理: 若\[n = \prod_{i = 1}^k = p_1^{a_1}\times p_2^{a_2} \times p_3^{a_3} \times \cdots \times p_k^{a_k}(其中p_i为n的质因数,a_i为正整数)\]
则\(n\)的所有正约数之和为\[f(n) = \prod_{i = 1}^k(\sum_{j = 0}^{a_i}p^j) \\=(1 + p_1 + p_1 ^ 2 + p_1^3+\cdots+ p_1^{a1}) \times(1 + p_2 + p_2^2+p_2^3+\cdots+p_2^{a_2})\times \dots \times (1 + p_k + p_k^2+p_k^3+\cdots + p_k^{a_k}\]

设\(S = 1 + p_k + p_k^2+p_k^3+\cdots + p_k^{a_k}\) 则\(p_kS = p_k + p_k^2+p_k^3+\cdots + p_k^{a_k}+p_k^{a_k+1}\)
后面的式子减前面的式子得:\((p_k - 1)S = p_k^{a_k+1} - 1\)即\[S = \frac{p_k^{a_k+1} - 1}{p_k - 1}\]
最后通过逆元+快速幂计算即可

#include<bits/stdc++.h>
#define gc getchar
#define ll long long
inline ll read(){ll x = 0; char ch = gc(); bool positive = 1;for (; !isdigit(ch); 
ch = gc()) if (ch == '-')  positive = 0;for (; isdigit(ch); ch = gc())  x = x * 10 
+ ch - '0';return positive ? x : -x;}inline void write(ll a){if(a>=10)write(a/10);
putchar('0'+a%10);}inline void writeln(ll a){if(a<0){a=-a; putchar('-');}write(a);
puts("");}

using namespace std;

const int N = 10000, p = 9901;

ll num[N], cnt[N], tot;

inline ll ksm(ll a, ll b) {
    ll res = 1;
    for(; b; b >>= 1, a = a * a % p)
        if(b & 1)
            res = res * a % p;
    return res;
}

int main() {
    ll x, a, b, y, ans = 1;
    x = read(), y = read();
    for(ll i = 2; i <= x; ++i)
        if(x % i == 0) {
            num[++tot] = i;
            while(x % i == 0) {
                ++cnt[tot];
                x /= i;
            }
            cnt[tot] *= y;
        }
    for(int i = 1; i <= tot; ++i) {
        a = ksm(num[i], cnt[i] + 1) - 1;
        b = ksm(num[i] - 1, p - 2);
        ans = ans * a * b % p;
    }
    writeln((ans + p ) % p);
    return 0;
}