Codeforces Round #338 (Div. 2) D. Multipliers 数论

D. Multipliers

题目连接：

http://codeforces.com/contest/615/problem/D

Description

Ayrat has number n, represented as it's prime factorization pi of size m, i.e. n = p1·p2·...·pm. Ayrat got secret information that that the product of all divisors of n taken modulo 109 + 7 is the password to the secret data base. Now he wants to calculate this value.

Input

The first line of the input contains a single integer m (1 ≤ m ≤ 200 000) — the number of primes in factorization of n.

The second line contains m primes numbers pi (2 ≤ pi ≤ 200 000).rst line of the input contains the string s — the coating that is present in the shop. Second line contains the string t — the coating Ayrat wants to obtain. Both strings are non-empty, consist of only small English letters and their length doesn't exceed 2100.

Output

Print one integer — the product of all divisors of n modulo 109 + 7.

Sample Input

2

2 3

Sample Output

36

Hint

题意

给你一个数的质因数，然后让你求出这个数所有因数的乘积

题解：

和hdu 5525很像，某场BC的原题

对于每个质因子，对答案的贡献为p^(d[p] * (d[p]-1) \ 2 * d[s])

d[p]表示p的因子数量，d[s]表示s这个数的因子数量

数量可以由因子数量定理求得,d[s] = (a1+1)(a2+1)...(an+1)，a1.a2.a3表示s的质因子的次数。

但是由于指数可能很大，所以我们就需要使用费马小定理就好了

但是又有除2的操作，mod-1有不是质数，不存在逆元，所以先对2（mod-1）取模。

代码

#include<bits/stdc++.h>
using namespace std;
#define maxn 200005
long long mod = 1e9+7;
long long mod2 = 2LL*(mod - 1);
long long quickpow(long long a,long long b,long long c)
{
    long long ans = 1;
    while(b)
    {
        if(b&1)ans = ans * a % c;
        a = a * a % c;
        b>>=1;
    }
    return ans;
}
int cnt[maxn];
int p[maxn];
int vis[maxn];
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&p[i]);
        cnt[p[i]]++;
    }
    long long tot = 1;
    for(int i=1;i<=n;i++)
    {
        if(vis[p[i]])continue;
        vis[p[i]]=1;
        tot = tot*(cnt[p[i]]+1)%mod2;//求因子数
    }
    memset(vis,0,sizeof(vis));
    long long ans = 1;
    for(int i=1;i<=n;i++)
    {
        if(vis[p[i]])continue;
        vis[p[i]]=1;
        ans=ans*quickpow(p[i],(tot*cnt[p[i]]/2)%mod2,mod)%mod;//每个数的贡献，费马小定理
    }
    cout<<ans<<endl;
}