#include<iostream>
#include<string>
#include<math.h>
using namespace std;
int main() {
int k; // j 的平方根
int result = 0;// 结果
int i = 0, num = 0;
cin >> num;
int j = 2,count=0;
int p1[10000];
for (j = 2; j <= num; j++) {
k = (int)sqrt((double)j);
for ( i = 2; i <= k; i++)
if (j % i == 0)
break;
if (i > k)
p1[count++] = j;
}
for (int z = 0; z < count; z++) {
if ((p1[z + 1] - p1[z]) == 2)
result++;
}
cout << result << endl;
return 0;
}
1007--乙级
最新推荐文章于 2024-10-24 21:29:43 发布