POJ 1862 Stripies(优先队列)

#pragma warning(disable:4996)#include <cstdio>#include <iostream>#include <queue>#include <cmath>using namespace std;//思路：//每次都去最大的两个碰撞，最后得到的一定最小//证明：//假设有a，b，c，答案是d，，那么d=2*sqrt(2*sqrt(a*b)*c)//d*d=8*sqrt(a*b*c*c)可知c的贡献比a和b要多//推断出，先碰撞的贡献少，后碰撞 的贡献多，，//所以要从大向小碰撞int main(){int n;while (scanf("%d", &n) != EOF){priority_queue<double>q;int tmp;for (int i = 1; i <= n; i++){scanf("%d", &tmp);q.push((double)tmp);}for (int i = 1; i <= n; i++){if (i == n){printf("%.3f\n", q.top());q.pop();break;}double a = q.top();q.pop();double b = q.top();q.pop();q.push(2.0*sqrt(a*b));}}return 0;}