【NYOJ】[169]素数

和素数距离问题基本类似
那个好像还比这个稍微难一些

不过鉴于是省赛的一道题目
所以还是再来写一遍吧

#include<stdio.h>
int a[1020]= {1,1};
int main() {
    for(int i=2; i<1020; i++) {
        if(!a[i]) {
            for(int j=i+i; j<1020; j+=i) {
                a[j]=1;
            }
        }
    }
    int T;
    scanf("%d",&T);
    while(T--) {
        int m;
        scanf("%d",&m);
        int t1=m,sum1=0;
        while(a[t1]) {
            sum1++;
            t1--;
        }
        int t2=m,sum2=0;
        while(a[t2]) {
            sum2++;
            t2++;
        }
        if(sum1<sum2)
            printf("%d\n",t1);
        else
            printf("%d\n",t2);
    }
    return 0;
}

标程竟然无耻的把1000以内的素数都直接打出来了……

#include<iostream>
#include<algorithm>
using namespace std;
int prime[]= {2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101,103,107,109,113,127,131,137,139,149,151,157,163,167,173,179,181,191,193,197,199,211,223,227,229,233,239,241,251,257,263,269,271,277,281,283,293,307,311,313,317,331,337,347,349,353,359,367,373,379,383,389,397,401,409,419,421,431,433,439,443,449,457,461,463,467,479,487,491,499,503,509,521,523,541,547,557,563,569,571,577,587,593,599,601,607,613,617,619,631,641,643,647,653,659,661,673,677,683,691,701,709,719,727,733,739,743,751,757,761,769,773,787,797,809,811,821,823,827,829,839,853,857,859,863,877,881,883,887,907,911,919,929,937,941,947,953,967,971,977,983,991,997,1009};
int main() {
    int n,m;
    cin>>n;
    while(n--) {
        cin>>m;
        if(m==1) {
            cout<<2<<endl;
            continue;
        }
        int* l=lower_bound(prime,prime+169,m);
        if(*l-m<=m-*(l-1)) cout<<*l<<endl;
        else cout<<*(l-1)<<endl;
    }
}

题目地址:【NYOJ】[169]素数