G - Non-Prime Factors

最新推荐文章于 2023-03-06 10:13:52 发布

king9666

最新推荐文章于 2023-03-06 10:13:52 发布

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博客聚焦编程竞赛中计算整数非质因数数量的问题。以整数100为例，说明非质因数的概念，给出输入输出格式，输入包含查询数量及具体整数，输出对应整数的非质因数数量，还提醒因I/O文件大需用快速I/O方法。

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G - Non-Prime Factors

In many programming competitions, we are asked to find (or count the number of) Prime Factors of an integer i. This is boring. This time, let’s count the number of Non-Prime Factors of an integer i, denoted as NPF(i).

For example, integer 100 has the following nine factors: {1,2⎯⎯,4,5⎯⎯,10,20,25,50,100}. The two which are underlined are prime factors of 100 and the rest are non-prime factors. Therefore, NPF(100) = 7

.
Input

The first line contains an integer Q
(1≤Q≤3⋅106) denoting the number of queries. Each of the next Q lines contains one integer i (2≤i≤2⋅106

).
Output

For each query i

, print the value of NPF(i).
Warning

The I/O files are large. Please use fast I/O methods.

Input

4
100
13
12
2018

Output

7
1
4
2

#include<bits/stdc++.h>
using namespace std;
bool vis[2000005];
int ans[2000005];
void init()
{

    vis[1]=1;
    int m=sqrt(2000002+0.5);
    for(int i=2; i<=m; ++i)
        if(!vis[i]) 
        for(int j=i*i; j<=2000002; j+=i) 
        vis[j]=1;
    for(int i = 1; i <= 2000000; ++i)
    {
        int rt = 2000000/i;
        for(int j = i; j <= rt; ++j)
        {
            if(vis[i])
            {
                ++ans[i*j];
            }
            if(vis[j] && i!=j)
            {
                ++ans[i*j];
            }
        }
    }
}
int main()
{
    init();
    int q;
    scanf("%d",&q);
    while(q--)
    {
        int n;
        scanf("%d",&n);
        printf("%d\n",ans[n]);
    }
    return 0;
}