Codeforces Round #338 (Div. 2) D. Multipliers（数学、降幂公式+逆元）

题意:

$给定1个大数的所有N\le 2×10^5个素因子, 求\prod 约数\ MOD\ 10^9 + 7$

分析:

$设这个数为X=p_1^{a_1}\cdot p_2^{a_2} \cdots p_n^{a_n}$
$考虑p_1^{a_1}对于约数积的贡献, 首先会在后面所有\prod_{i=2}^n(a_i+1)个约数中出现$
$且p_1^{a_1}取值有p_1^0、p_1^1\cdots p_1^{a_1}, 那么p_1的贡献:$
$E_{p_1}=(p_1^0\cdot p_1^1\cdots p_1^{a_1})^{\prod_{i=2}^n(a_i+1)}$
　　 $=(p_1^{\frac{a_1\cdot (a_1+1)}{2}})^{\prod_{i=2}^n(a_i+1)}$
　　 $=p_1^{\frac{a_1\cdot \prod_{i=1}^n(a_i+1)}{2}}$
$同理E_{p_i}=p_i^{\frac{a_i\cdot \prod_{i=1}^n(a_i+1)}{2}}$

---

$由于指数非常大, 可以根据a^{p-1}\equiv1\ mod\ p(Fermat\ Little\ Theorem)来降幂, 即对指数p-1取模$
$或者降幂公式也可以, a^x\equiv a^{x\%\phi(C)+\phi(C)}\ mod\ C(x\ge \phi(C))$
$由于有除法, 需要考虑逆元,a/b\ mod\ m = a\ mod\ bm/b$
$证明如下:$
$设a/b=km+r, 则a=kbm+br$
$右边=(kbm+br)\ mod \ bm/b=br/b=r=左边　\blacksquare$

代码:

//
//  Created by TaoSama on 2016-01-09
//  Copyright (c) 2015 TaoSama. All rights reserved.
//
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>

using namespace std;
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
const int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;

typedef long long LL;

int n;

LL ksm(LL x, LL n) {
    LL ret = 1;
    for(; n; n >>= 1) {
        if(n & 1) ret = ret * x % MOD;
        x = x * x % MOD;
    }
    return ret;
}

int main() {
#ifdef LOCAL
    freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
    ios_base::sync_with_stdio(0);

    while(scanf("%d", &n) == 1) {
        map<int, int> cnt;
        for(int i = 1; i <= n; ++i) {
            int x; scanf("%d", &x);
            ++cnt[x];
        }

        LL E = 1, M = 2 * (MOD - 1);
        for(auto &x : cnt) E = E * (x.second + 1) % M;

        LL ans = 1;
        for(auto &x : cnt) {
            LL e = x.second * E % M / 2;
            ans = ans * ksm(x.first, e) % MOD;
        }
        printf("%I64d\n", ans);
    }
    return 0;
}