csu 1801(合数分解+排列组合)

1801: Mr. S’s Romance

Time Limit: 1 Sec  Memory Limit: 128 MB
Submit: 15  Solved: 5
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Description

Mr. S is a young math professor who is famous for being crazy about collecting prime numbers. Once he met a number, he would first divide it into several prime numbers and then push them into his magic box. Every week, Mr. S  sell out the primes in his box to earn money. 

 

Today is August 22th, which is an important day to Mr. S because it’s his girl friend’s birthday. She is a perfect girl, so Mr. S decide to choose some numbers from his magic box to multiply to form a perfect square number as a love present.Undoubtedly, he must pick up one number.

This week, Mr. S has totally met n numbers a1,a2,...,an. Suddenly, Mr.S come up with an exciting question that how many different ways of choices can form a perfect square number as a present. Can you help him solve it? The answer maybe too large, so you should output the answer modulo by 1000000007.

Input

First line is a positive integer T (<=5), represents there are T test cases.

For each test case:

First line includes a number n(1≤n≤100000)，next line there are n numbers a1,a2,...,an,(1<ai≤10^6).

Output

For the i-th test case , first output Case #i: in a single line.

Then output the answer of i-th test case modulo by 1000000007.

Sample Input

2
3
3 3 4
3
2 2 2

Sample Output

Case #1:
3
Case #2:
3


题意:在 n 个整数的质因子的里面选若干个组成完全平方数的种类数?
题解:把各个数分解得到质因子数 ,然后要是完全平方数,那么每种质因子的个数都要是偶数,对于某个质因子假设个数为 x,那么选法就有 C(x,0)+C(x,2)+..+C(x,最接近x的那个偶数) = 2^(x-1) 通过排列组合得解.

答案要减掉全部都不选的那种.

#include <iostream>
#include <cstring>
#include <stdio.h>
#include <stdlib.h>
#include <algorithm>
using namespace std;
typedef long long LL;
const LL mod = 1000000007;
const int N = 1000005;
int n;
LL prime[N+1];
LL num[N];
void getPrime()
{
    memset(prime,0,sizeof(prime));
    for(int i=2; i<=N; i++)
    {
        if(!prime[i])prime[++prime[0]]=i;
        for(int j=1; j<=prime[0]&&prime[j]<=N/i; j++)
        {
            prime[prime[j]*i]=1;
            if(i%prime[j]==0) break;
        }
    }
}
LL fatCnt;
void getFactors(LL x)
{
    LL tmp=x;
    for(int i=1; prime[i]<=tmp/prime[i]; i++)
    {
        if(tmp%prime[i]==0)
        {
            while(tmp%prime[i]==0)
            {
                num[prime[i]]++;
                tmp/=prime[i];
            }
        }
    }
    if(tmp!=1) num[tmp]++;
}
LL pow_mod(LL a,LL n){
    LL ans = 1;
    while(n){
        if(n&1) ans = ans*a%mod;
        a = a*a%mod;
        n>>=1;
    }
    return ans;
}
int main()
{
    int tcase,t=1;
    scanf("%d",&tcase);
    getPrime();
    while(tcase--)
    {
        memset(num,0,sizeof(num));
        int n;
        scanf("%d",&n);
        LL a;
        for(int i=0; i<n; i++)
        {
            scanf("%lld",&a);
            getFactors(a);
        }
        LL ans = 1;
        for(int i=1;i<N;i++){
            if(num[i]>1){
                ans = ans*pow_mod(2,num[i]-1)%mod;
            }
        }
        printf("Case #%d:\n%lld\n",t++,ans-1);
    }
    return 0;
}