SGU 113素数筛

最新推荐文章于 2020-03-30 19:51:31 发布

原创最新推荐文章于 2020-03-30 19:51:31 发布 · 964 阅读

0 ·

CC 4.0 BY-SA版权

ACM_水题专栏收录该内容

22 篇文章

订阅专栏

113. Nearly prime numbers

time limit per test: 0.5 sec.
memory limit per test: 4096 KB

Nearly prime number is an integer positive number for which it is possible to find such primes P₁ and P₂ that given number is equal to P₁*P₂. There is given a sequence on N integer positive numbers, you are to write a program that prints “Yes” if given number is nearly prime and “No” otherwise.

Input

Input file consists of N+1 numbers. First is positive integer N (1£N£10). Next N numbers followed by N. Each number is not greater than 10⁹. All numbers separated by whitespace(s).

Output

Write a line in output file for each number of given sequence. Write “Yes” in it if given number is nearly prime and “No” in other case.

Sample Input

1
6

Sample Output

Yes

Author	: Michael R. Mirzayanov
Resource	: PhTL #1 Training Contests
Date	: Fall 2001

一个数如果能被写成两个素数之积称之为near prime。

/**********************
* author:crazy_石头
* Pro:SGU 113
* algorithm:素数筛
* Time:15ms
* Judge Status:Accepted
***********************/
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <queue>
#include <vector>
#include <algorithm>

using namespace std;

#define rep(i,h,n) for(int i=(h);i<=(n);i++)
#define ms(a,b) memset((a),(b),sizeof(a))
#define eps 1e-6
#define INF 1<<29
#define LL __int64
const int maxn=40000+5;
const int maxm=200+10;

int prime[maxn],cnt;
int p[maxn];
int n;

inline void init_prime()
{
    ms(p,0);
    p[0]=p[1]=1;//0和1不是素数；
    rep(i,2,maxn)
    {
        if(!p[i])
        {
            for(int j=i+i;j<=maxn;j+=i)
            p[j]=1;//若i是素数，则它的倍数不是素数，改为1；
        }
    }
    rep(i,2,maxn)
    {
        if(!p[i])
        prime[cnt++]=i;
    }
}

inline bool judge(int n)
{
    if(n<2)return false;
    for(int i=2;i*i<=n;i++)
        if(n%i==0)return false;
    return true;
}

int main()
{
    int test;
    scanf("%d",&test);
    while(test--)
    {
        cnt=0;
        init_prime();
        int ok=0;
        scanf("%d",&n);
        rep(i,0,cnt-1)
        {
            if(n%prime[i]==0&&judge(n/prime[i]))
            {//n等于两个素数之积，说明n%prime[i]==0并且另一个除数也是素数；
                ok=1;
                break;
            }
        }
        if(ok)
            printf("Yes\n");
        else
            printf("No\n");
    }
    return 0;
}