51nod 1106 质数检测 （三种模板_好题）

Input

第1行：一个数N，表示正整数的数量。(1 <= N <= 1000)
第2 - N + 1行：每行1个数(2 <= S[i] <= 10^9)

Output

输出共N行，每行为 Yes 或 No。

Input示例

5
2
3
4
5
6

Output示例

Yes
Yes
No
Yes
No

效率最低：

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
bool is_prime(int num)
{
	int i;
	if(num==1) return false;
	if(num==2) return true;
	for(i=2;i*i<=num;i++) {
		if(num%i==0) return false;
	}
	return true;
}
int main()
{
	int n,num;
	scanf("%d",&n);
	while(n--) {
		scanf("%d",&num);
		if(is_prime(num)) printf("Yes\n");
		else printf("No\n");
	}
	return 0;
}

素数测试:

#include<cstdlib>
#include<cstdio>
int modularExponent(int a, int b, int n) {
	int ret = 1;
	for (; b; b >>= 1, a = (int) ((long long) a * a % n)) {
		if (b & 1) {
			ret = (int) ((long long) ret * a % n);
		}
	}
	return ret;
}
bool millerRabin(int n,int a) {
	if (n == 1 || (n != 2 && !(n % 2)) || (n != 3 && !(n % 3)) || (n != 5 && !(n % 5)) || (n != 7 && !(n % 7))) {
		return false;
	}
	int r = 0, s = n - 1, j;
	if(!(n%a)) return false;
	while(!(s&1)){ s >>= 1; r++; }
	long long k = modularExponent(a, s, n);
	if(k == 1) return true;
	for(j = 0; j < r; j++, k = k * k % n)
	    if(k == n - 1) return true;
	return false;
}
bool miller_Rabin(int n)//
{
	int a[]={2,3,5,7},i;//能通过测试的最小素数为 3215031751(此数超int)
	for(i=0;i<4;i++){
		if(!millerRabin(n,a[i]))return false;
	}
	return true;
}
int main()
{
	int n,x;
	scanf("%d",&n);
	while(n--){
		scanf("%d",&x);
		printf("%s\n",miller_Rabin(x)?"Yes":"No");
	}
	return 0;
}

神人写的：

#include <stdio.h>
#include <math.h>

#define  MAXP 31627

char  flag[MAXP+1];
int   prime[4000];
int   count = 0;

void  init_ptbl() {
        int  i, k, m;
        prime[count++] = 2;
        for( i = 3; i <= MAXP / 2; i += 2 ) {
                if( flag[i] == 1 ) continue;
                for( k = 2; ( m = i * k ) <= MAXP; ++k ) flag[m] = 1;
        }
        for( i = 3; i <= MAXP; i += 2 ) if( flag[i] == 0 ) prime[count++] = i;
}

int main() {
        int n, x, i, e;
        init_ptbl();

  scanf( "%d", &n );
  while( n-- ) {
    scanf( "%d", &x );
    if( ( x & 1 ) == 0 ) { puts( x == 2 ? "Yes" : "No" ); continue; }
    e = (int)sqrt( x ) + 1;
    for( i = 1; prime[i] <= e; ++i ) if( x % prime[i] == 0 ) break;
    puts( prime[i] <= e ? "No" : "Yes" );
  }
  return 0;
}