codeforces 893e Counting Arrays

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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E. Counting Arrays
time limit per test
3 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

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;
  • .

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);  
    }  
}  

### Codeforces 887E Problem Solution and Discussion The problem **887E - The Great Game** on Codeforces involves a strategic game between two players who take turns to perform operations under specific rules. To tackle this challenge effectively, understanding both dynamic programming (DP) techniques and bitwise manipulation is crucial. #### Dynamic Programming Approach One effective method to approach this problem utilizes DP with memoization. By defining `dp[i][j]` as the optimal result when starting from state `(i,j)` where `i` represents current position and `j` indicates some status flag related to previous moves: ```cpp #include <bits/stdc++.h> using namespace std; const int MAXN = ...; // Define based on constraints int dp[MAXN][2]; // Function to calculate minimum steps using top-down DP int minSteps(int pos, bool prevMoveType) { if (pos >= N) return 0; if (dp[pos][prevMoveType] != -1) return dp[pos][prevMoveType]; int res = INT_MAX; // Try all possible next positions and update 'res' for (...) { /* Logic here */ } dp[pos][prevMoveType] = res; return res; } ``` This code snippet outlines how one might structure a solution involving recursive calls combined with caching results through an array named `dp`. #### Bitwise Operations Insight Another critical aspect lies within efficiently handling large integers via bitwise operators instead of arithmetic ones whenever applicable. This optimization can significantly reduce computation time especially given tight limits often found in competitive coding challenges like those hosted by platforms such as Codeforces[^1]. For detailed discussions about similar problems or more insights into solving strategies specifically tailored towards contest preparation, visiting forums dedicated to algorithmic contests would be beneficial. Websites associated directly with Codeforces offer rich resources including editorials written after each round which provide comprehensive explanations alongside alternative approaches taken by successful contestants during live events. --related questions-- 1. What are common pitfalls encountered while implementing dynamic programming solutions? 2. How does bit manipulation improve performance in algorithms dealing with integer values? 3. Can you recommend any online communities focused on discussing competitive programming tactics? 4. Are there particular patterns that frequently appear across different levels of difficulty within Codeforces contests?
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