|Tyvj|P1001 第K极值

http://tyvj.cn/p/1001

排序后找到两个数相减，再判断一下质数即可

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int a[10005];int main () {int n,k;scanf("%d%d", &n, &k);for (int i=0;i<n;i++) scanf("%d", &a[i]);sort(a,a+n);int ans = a[n-k] - a[k-1];if (ans<2) {printf("NO\n%d\n", ans);return 0;} else{for (int i=2;i<=ans/2;i++){if (ans % i==0) {printf("NO\n%d\n", ans);return 0;}}printf("YES\n%d\n", ans);}return 0;}