SGU113

定义一个数x是nearly prime当且仅当x由两个素数相乘得到，给定n个数，判断他们是不是nearly prime

傻×题

其实可以不用筛的，我脑残一下筛了素数

//Lib#include<cstdio>#include<cstring>#include<cstdlib>#include<cmath>#include<ctime>#include<iostream>#include<algorithm>#include<vector>#include<string>#include<queue>using namespace std;//Macro#define rep(i,a,b) for(int i=a,tt=b;i<=tt;++i)#define rrep(i,a,b) for(int i=a,tt=b;i>=tt;--i)#define erep(i,e,x) for(int i=x;i;i=e[i].next)#define irep(i,x) for(__typedef(x.begin()) i=x.begin();i!=x.end();i++)#define read() (strtol(ipos,&ipos,10))#define sqr(x) ((x)*(x))#define pb push_back#define PS system("pause");typedef long long ll;typedef pair<int,int> pii;const int oo=~0U>>1;const double inf=1e20;const double eps=1e-6;string name="",in=".in",out=".out";//Varconst int maxn=40000;int vis[40008],p[10008],cnt,T,n;void Pre(){rep(i,2,maxn){if(!vis[i])p[++cnt]=i;for(int j=1;j<=cnt&&i*p[j]<=maxn;j++){vis[i*p[j]]=p[j];if(i%p[j]==0)break;}}}bool isPrime(int n){if(n<2)return false;rep(i,2,sqrt((double)n)){if(n%i==0)return false;}return true;}void Work(){scanf("%d",&T);bool f;rep(i,1,T){scanf("%d",&n);f=false;rep(i,1,cnt){if(n%p[i]==0){if(isPrime(n/p[i]))f=true;break;}}cout<<(f?"Yes":"No")<<endl;}}int main(){//freopen((name+in).c_str(),"r",stdin);//freopen((name+out).c_str(),"w",stdout);//Init();Pre();Work();return 0;}