HDU 5525 Product

Problem Description

A1,A2....An,indicating  N=∏ni=1iAi.What is the product of all the divisors of N?

 

Input

There are multiple test cases.
First line of each case contains a single integer n. (1≤n≤105)
Next line contains n integers  A1,A2....An,it's guaranteed not all  Ai=0. (0≤Ai≤105).
It's guaranteed that  ∑n≤500000.

 

Output

109+7

 

Sample Input

4 0 1 1 0 5 1 2 3 4 5

 

Sample Output

36 473272463

要注意过程当中的溢出问题

#include<cstdio>
#include<cmath>
#include<queue>
#include<stack>
#include<cstring>
#include<vector>
#include<algorithm>
using namespace std;
typedef long long LL;
const LL base = 1e9 + 7;
const LL mod = 1e9 + 6;
const int maxn = 100005;
int n, a[maxn], f[maxn], p[maxn], tot = 0, x;
LL ans[maxn], qur, L[maxn], R[maxn];

struct point
{
  int x, y;
  point(int x = 0, int y = 0) :x(x), y(y){}
};
vector<point> map[maxn];

void pre()
{
  for (int i = 2; i < maxn; i++)
  {
    if (!f[i]) { map[i].push_back(point(tot, 1)); f[i] = tot; p[tot++] = i; }
    else
    {
      for (int j = 0; j < tot; j++)
      if (i%p[j] == 0)
      {
        int u = i / p[j], v = 1;
        for (int k = 0; k < map[u].size(); k++)
          if (map[u][k].x != j) map[i].push_back(map[u][k]);
          else v += map[u][k].y;
        map[i].push_back(point(j, v));
        break;
      }
    }
    for (int j = 0; j < tot&&i*p[j] < maxn; j++)
    {
      f[i*p[j]] = -1;
      if (i%p[j] == 0) break;
    }
  }
}

LL get(LL x, LL y)
{
  LL ans = 1;
  for (x %= base; y; y >>= 1)
  {
    if (y & 1) ans = ans*x%base;
    x = x*x%base;
  }
  return ans;
}

int main()
{
  pre();
  while (scanf("%d", &n) != EOF)
  {
    for (int i = 0; i <= tot; i++) ans[i] = 0;
    for (int i = 1; i <= n; i++)
    {
      scanf("%d", &x);
      for (int j = 0; j < map[i].size(); j++)
      {
        ans[map[i][j].x] += (LL)map[i][j].y*x;
      }
    }
    for (int i = 0; i < tot; i++)
    {
      L[i] = (ans[i] + 1) % mod;
      if (i) L[i] = L[i] * L[i - 1] % mod;
    }
    for (int i = tot; i; i--)
    {
      R[i] = (ans[i] + 1) % mod;
      if (i < tot) R[i] = R[i] * R[i + 1] % mod;
    }
    qur = 1;
    for (int i = 0; i < tot; i++)
    {
      LL u = ans[i], v = ans[i] + 1;
      if (u % 2 == 0) u = u / 2; else v = v / 2;
      u = (u%mod)*(v%mod) % mod;
      if (i) u = u*L[i - 1] % mod;
      u = u*R[i + 1] % mod;
      qur = qur*get(p[i], u) % base;
    }
    printf("%I64d\n", qur);
  }
  return 0;
}