@(K ACMer)

題意:
給你一個n個數的數列:an. 求

分析:
知識補充:
- 一個數的約數的個數等於:

回到這個題,我們先把所有的質因子和他的次數算出來,然後每一個質因子被利用的次數等於其它不含該因子的質數所能構成的約數的個數乘以:(1+2+3+....+n

#include <iostream>
#include <cstdio>
#include <cstring>
#include <set>
#include <map>
#include <stack>
#include <vector>
#include <string>
#include <queue>
#include <cstdlib>
 

#include <cmath>
#include <algorithm>
using namespace std;
typedef pair<int, int> pii;
typedef long long ull;
typedef long long ll;
typedef vector<int> vi;
#define xx first
#define yy second
#define rep(i, a, n) for (int i = a; i < n; i++)
#define sa(n) scanf("%d", &(n))
 

#define vep(c) for(decltype((c).begin()) it = (c).begin(); it != (c).end(); it++)
const int mod = int(1e9) + 7, INF = 0x3fffffff, maxn = 1e5 + 12;
int n, a[maxn], b[maxn], prim[maxn / 10], cnt;
ll sum, t;
ll mp[maxn], suffix[maxn / 10], prefix[maxn / 10];

void getit(void)
{
    memset(b, 1, sizeof(b));
    rep (i, 2, maxn) {
        if (b[i]) {
            prim[cnt++] = i;
            for (int j = i; j < maxn; j += i)
                b[j] = 0;
        }
    }
}

int pows(ll y, ll n)
{
    ll ret = 1;
    while (n) {
        if (n & 1) ret = ret * y % mod;
        y = y * y % mod;
        n >>= 1;
    }
    return int(ret % mod);
}

void doit(int x, int y)
{
    if (y == 0) return;
    for (int i = 0; prim[i] * prim[i] <= x; i++) {
        if (x % prim[i] == 0) {
            ll tt = 0;
            while (x % prim[i] == 0) {
                tt += y;
                x /= prim[i];
            }
            mp[prim[i]] += tt;
        }
    }
    if (x != 1) mp[x] += y;
}

int mul(ll x) {
    ll a = x, b = x + 1;
    if (a % 2) b /= 2;
    else a /= 2;
    return (a % (mod - 1)) * (b % (mod - 1)) % (mod - 1);
}

int main(void)
{
    getit();
    while (~sa(n)) {
        rep (i, 0, n) {
            sa(a[i]);
        }
        memset(mp, 0, sizeof(mp));
        sum = 1, t = 1;
        rep (i, 2, n + 1) {
            doit(i, a[i - 1]);
        }
        suffix[cnt - 1] = prefix[0] = 1;
        rep (i, 1, cnt) prefix[i] = prefix[i - 1] * (mp[prim[i - 1]] + 1) % (mod - 1);
        for (int i = cnt - 2; i >= 0; i--) suffix[i] = suffix[i + 1] * (mp[prim[i + 1]] + 1) % (mod - 1);
        ll ans = 1;
        rep (i, 0, cnt) {
            ll temp = pows(prim[i], (prefix[i] * suffix[i] % (mod - 1)) * mul(mp[prim[i]]) % (mod - 1));
            ans = ans * temp % mod;
        }
        printf("%lld\n", ans);
    }
    return 0;
}