[51nod 1106] 质数检测

给出N个正整数，检测每个数是否为质数。如果是，输出"Yes"，否则输出"No"。

(N leq 1000, a_i leq 10^9)

Solution

先筛出 (10^5) 以内的质数，利用这些质数去判断每个给定的数是否为质数，复杂度 (O(N sqrt V/ log V))

#include <bits/stdc++.h>
using namespace std;

const int MAXN = 100005;
int prime[MAXN+1]; // Note: Let prime[0] donate the number of primes
// Note: the array "prime" has two different roles in the algorithm
void presolve() {
    memset(prime,0,sizeof prime);
    for(int i=2;i<=MAXN;i++) {
        if(!prime[i]) prime[++prime[0]]=i;
        for(int j=1;j<=prime[0]&&prime[j]<=MAXN/i;j++) {
            prime[prime[j]*i]=1;
            if(i%prime[j]==0) break;
        }
    }
}

signed main() {
    presolve();
    int n,t;
    cin>>n;
    while(n--) {
        cin>>t;
        int flag=0;
        int lim=sqrt(t);
        for(int i=1;prime[i]<=lim;i++) flag|=(t%prime[i]==0);
        cout<<(flag?"No":"Yes")<<endl;
    }
}