Input
Input file consists of N+1 numbers. First is positive integerN (1£N£10). NextN numbers followed byN. 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
先來複習一下篩法求素數相關知識:
#include<iostream>
#include<string.h>
#include<stdio.h>
#include<ctype.h>
#include<algorithm>
#include<stack>
#include<queue>
#include<set>
#include<math.h>
#include<vector>
#include<map>
#include<deque>
#include<list>
using namespace std;
int vis[9999];
void isprime(int n)
{
int k=sqrt(n+0.5);
for(int i=2; i<=k; i++)
{
if(!vis[i])
for(int j=i*i; j<=n; j+=i)
vis[j]=1;
}
}
int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
memset(vis,0,sizeof(vis));
vis[0]=1,vis[1]=1;
isprime(n);
for(int i=0; i<=n; i++)
{
if(!vis[i])
printf("%d ",i);
}
}
return 0;
}
對於給定的數據很大、需用篩法素數打表。
那篩法求素數對該題目有什麼作用呢?沒啥關係。。只不過看到素數就想到了篩法求素數 ^-^
上述題目題意爲給出一個數,讓你判斷這個數是否爲Nearly prime number,Nearly prime number的定義是,這個數能被2個素數相乘求得。
#include<iostream>
#include<string.h>
#include<stdio.h>
#include<ctype.h>
#include<algorithm>
#include<stack>
#include<queue>
#include<set>
#include<math.h>
#include<vector>
#include<map>
#include<deque>
#include<list>
using namespace std;
bool prime(int p)
{
int w=0;
int k=sqrt(p*1.0);
for(int i=2; i<=k; i++)
if(p%i==0)
{
w=1;
break;
}
if(w==1)
return 0;
else
return 1;
}
bool f(int n)
{
int f=0;
for(int i=2; i<=sqrt(n*1.0); i++)
{
if(prime(i))
{
if(n%i==0)//確保n/i爲整數(能被整除)
{
if(prime(n/i))
{
f=1;
break;
}
}
}
}
if(f==1)
return true;
else
return false;
}
int main()
{
int t,n;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
if(f(n))
printf("Yes\n");
else
printf("No\n");
}
return 0;
}