CC NUMFACT(Number of Factors-枚举因子)

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Number of Factors

All submissions for this problem are available.

Alice has learnt factorization recently. Bob doesn't think she has learnt it properly and hence he has decided to quiz her. Bob gives Alice a very large number and asks her to find out the number of factors of that number. To make it a little easier for her, he represents the number as a product of N numbers. Alice is frightened of big numbers and hence is asking you for help. Your task is simple. Given Nnumbers, you need to tell the number of distinct factors of the product of these N numbers.

Input:

First line of input contains a single integer T, the number of test cases.

Each test starts with a line containing a single integer N.
The next line consists of N space separated integers (A_i).

Output:

For each test case, output on a separate line the total number of factors of the product of given numbers.

Constraints:

1 ≤ T ≤ 100
1 ≤ N ≤ 10
2 ≤ A_i ≤ 1000000

Example:

Input:

Output:

8
10
3

Scoring:

You will be awarded 40 points for correctly solving for A_i ≤ 100.

You will be awarded another 30 points for correctly solving for A_i ≤ 10000.

The remaining 30 points will be awarded for correctly solving for A_i ≤ 1000000.

暴力求出所有因子，枚举取的个数+1相乘

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (1000000007)
#define MP make_pair
#define MAXN (1000000+1)
#define MAXAi (1000000)
long long mul(long long a,long long b){return (a*b)%F;}
long long add(long long a,long long b){return (a+b)%F;}
long long sub(long long a,long long b){return (a-b+(a-b)/F*F+F)%F;}
typedef long long ll;
int n,a[MAXN]={0};
int main()
{
	freopen("NUMFACT.in","r",stdin);
//	freopen(".out","w",stdout);
    int T;
    scanf("%d",&T);
    while (T--)
    {
        memset(a,0,sizeof(a));
        scanf("%d",&n);
        For(i,n)
        {
            int p;
            scanf("%d",&p);
            Fork(i,2,sqrt(p)) while (p%i==0) a[i]++,p/=i;
            if (p>1) a[p]++;
        }
        ll ans=1;
        For(i,MAXAi)
        {
            if (a[i]) ans*=a[i]+1;
        }
        cout<<ans<<endl;


    }
	return 0;
}