Non-Prime Factors（素数筛 + 质因分解）
Within numerous programming contests, participants are typically required to identify (or enumerate) the Prime Factors associated with a given integer ii. This task is rather monotonous. Conversely, this iteration involves counting the Non-Prime Factors composing a specific integer ii and representing this quantity as NPF(i).
For instance, number 1₀₀₁₀₀ is characterized by nine factors: {1,,2,,,4,,5,,, etc.} The specific elements highlighted correspond to its prime factorization while the others represent composite factorization. Consequently,NPF(₁₀₀) equals ₇₇.
Input
The first line includes an integer value QQ, where 1 ≤ Q ≤ 3×10^6, representing the number of queries. For each subsequent line in these QQ lines, there is an integer ii (where 2 ≤ i ≤ 2×10^6) present.
Output
For each query ii, print the value of NPF(i).
Warning
The I/O files are large. Please use fast I/O methods.
| Sample Input 1 | Sample Output 1 |
|---|---|
4
100
13
12
2018||
7
1
4
2||
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#define maxx 2e6+5
int vis[2000002];
int ans[2000002];
void f()
{
vis[1]=1;
for(int i=2; i<=sqrt(maxx); ++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()
{
int t;
f();
scanf("%d", &t);
while(t--)
{
int n;
scanf("%d", &n);
printf("%d\n", ans[n]);
}
return 0;
}