Codeforces Round #311 (Div. 2)--Pasha and Tea--二分
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原题链接:https://acm.bnu.edu.cn/v3/problem_show.php?pid=49673
题意:一壶茶w毫升,n个男,n个女,要求男生分到的茶都一样多;女生一样。但是每个男生分到的毫升是每个女生的两倍,求w的最大值。
分析:首先排序,分为两组,1–n,n+1–2*n,很容易知道前面是女生的,后面是男生的。设一个女生分到x毫升,则有n*x+n*(2*x)<=w,求得x<=w/(3*n)。最后二分[ 0 , min(x , a[ 1 ]) ]。
AC代码:
#define _CRT_SECURE_NO_DEPRECATE#include<iostream>#include<vector>#include<cstring>#include<queue>#include<stack>#include<algorithm>#include<cmath>#include<string>#include<stdio.h>#define INF 1000000000#define EPS 1e-6using namespace std;int n, w;double a[200005];int main(){ while (~scanf("%d%d", &n, &w)) { for (int i = 1; i <= 2 * n; i++) scanf("%lf", &a[i]); sort(a + 1, a + 1 + 2 * n); double l = 0.0; double r = min(a[1], (double)w / 3 * n); while (r - l >= EPS) { double mid = (l + r) / 2; if (mid * 2 < a[n + 1]) l = mid; else if (mid * 2 > a[n + 1]) r = mid; else { l = mid; break; } } printf("%lf\n", 3 * n* l); } return 0;}
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