codeforces 893e Counting Arrays

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最新推荐文章于 2022-09-22 22:02:28 发布

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本文介绍了一个算法问题，即计算特定长度的整数数组，其元素乘积等于给定整数x的数量。通过质因数分解将问题简化，并利用组合数学中的放球问题计算方案数，最终得出解法。

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You are given two positive integer numbers x and y. An array F is called an y-factorization of x iff the following conditions are met:

There are y elements in F, and all of them are integer numbers;
$\text{[math]}$ .

You have to count the number of pairwise distinct arrays that are y-factorizations of x. Two arrays A and B are considered different iff there exists at least one index i (1 ≤ i ≤ y) such that Ai ≠ Bi. Since the answer can be very large, print it modulo 109 + 7.

Input

The first line contains one integer q (1 ≤ q ≤ 105) — the number of testcases to solve.

Then q lines follow, each containing two integers xi and yi (1 ≤ xi, yi ≤ 106). Each of these lines represents a testcase.

Output

Print q integers. i-th integer has to be equal to the number of yi-factorizations of xi modulo 109 + 7.

   Example 
 

     input 
    
      Copy 
    
2
6 3
4 2

     output 
   
36
6

Note

In the second testcase of the example there are six y-factorizations:

{ - 4, - 1};
{ - 2, - 2};
{ - 1, - 4};
{1, 4};
{2, 2};

{4, 1}.

题意：求长度为y的整数数组，它的所有数的乘积为x的数组的个数。

题解：

首先我们对x进行质因数分解，这样就转化成了若干个独立的小问题。

之后我们发现其实就是把x的质因子分配到y数组的数字上的方案数相乘，放球问题，C（y-1+k，y-1）。

因为可以为负数，所以最后答案乘上C（y,0）*C（y,2）*C(y,4)*......

这个数字是2^(y-1)

代码：

#include <bits/stdc++.h>  
using namespace std;  
typedef long long ll;  
ll q,a[30],x,y,fac[2000005],two[2000005];  
ll qmod(ll n,ll k)  
{  
    ll ans=1;  
    while(k) 
    {  
        if(k&1) ans=ans*n%1000000007;  
        k>>=1;  
        n=n*n%1000000007;  
    }  
    return ans%1000000007;  
}  
ll inv(ll n){return qmod(n,1000000005);}  
ll C(ll m,ll n)  
{  
    ll M=fac[m];  
    ll NM=fac[n]*fac[m-n]%1000000007;  
    NM=M*inv(NM)%1000000007;  
    return NM;  
}
int main()  
{  
    fac[0]=two[0]=1;  
    for(int i=1;i<2000005;i++) 
    {  
        two[i]=two[i-1]*2LL%1000000007;  
        fac[i]=fac[i-1]*i%1000000007;  
    }  
    cin>>q;
    while(q--) 
    {  
        memset(a,0,sizeof(a));  
        int tot=0;  
        ll ans=1;  
        scanf("%I64d%I64d",&x,&y);  
        for(int i=2;i*i<=x;i++) 
            if(x%i==0) 
                for(tot++;x%i==0;x/=i) 
                    a[tot]++;
        if(x!=1) a[++tot]=1;  
        for(int i=1;i<=tot;i++) ans=ans*C(y+a[i]-1,a[i])%1000000007;  
        ans=ans*two[y-1]%1000000007;  
        printf("%I64d\n",ans);  
    }  
}