标签:its word code put eof can rmi inpu art
1 2 3 4 5 17 0
1: no 2: no 3: yes 4: no 5: yes 6: yes
AC代码:
#include <map>
#include <set>
#include <cmath>
#include <deque>
#include <queue>
#include <stack>
#include <cstdio>
#include <cctype>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#define LL long long
#define INF 0x7fffffff
using namespace std;
const int maxn = 20000;
int prime[20000];
void init() {
prime[1] = 2;
for(int i = 2; i < maxn; i ++)
if(prime[i] == 0) {
prime[i] = 1;
for(int j = i * 2; j < maxn; j += i) {
prime[j] = 2;
}
}
prime[2] = 2;
}
int main() {
init();
int cas = 1;
int a;
while(scanf("%d", &a) != EOF) {
if(a == 0) break;
if(prime[a] == 1) {
printf("%d: yes\n", cas ++);
}
else if(prime[a] == 2) {
printf("%d: no\n", cas ++);
}
}
return 0;
}
标签:its word code put eof can rmi inpu art
原文地址:http://www.cnblogs.com/jzdwajue/p/7101893.html