HDU - 2161 - Primes （质数）

Posted 2020-09-26 jzdwajue

Primes

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8959    Accepted Submission(s): 3754

Problem Description

Write a program to read in a list of integers and determine whether or not each number is prime. A number, n, is prime if its only divisors are 1 and n. For this problem, the numbers 1 and 2 are not considered primes. 

 

Input

Each input line contains a single integer. The list of integers is terminated with a number<= 0. You may assume that the input contains at most 250 numbers and each number is less than or equal to 16000.

 

Output

The output should consists of one line for every number, where each line first lists the problem number, followed by a colon and space, followed by "yes" or "no". 

 

Sample Input

1
2
3
4
5
17
0

 

Sample Output

1: no
2: no
3: yes
4: no
5: yes
6: yes

 

Source

2008 “Sunline Cup” National Invitational Contest - Warm Up

 

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;
}