HDU1969 PIE 二分查找

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题目链接： http://acm.hdu.edu.cn/showproblem.php?pid=1969

题目意思：有n个块饼和M+1个人要分饼，且每人一块，饼不能拼接，要求每人分得的最大的饼的面积
思路：二分，先用总的饼的面积 / 人数 = 理论上最大面积，取为上界high，再求一个mid ，一直到循环退出，不能在中途找到时结束循环

if(饼能分成M+1份)

{ left = mid }

else { right = mid }

#include<cstring>#include<string>#include<fstream>#include<iostream>#include<iomanip>#include<cstdio>#include<cctype>#include<algorithm>#include<queue>#include<map>#include<set>#include<vector>#include<stack>#include<ctime>#include<cstdlib>#include<functional>#include<cmath>using namespace std;#define PI acos(-1.0)#define MAXN 100005#define eps 1e-7#define INF 0x7FFFFFFF#define ff sqrt(5.0)double val[100005];int n,m,t;int check(double x){    int sum=0;    for(int i = 0;i < n;i++)    {        sum += val[i] / x;    }    if(sum >= m + 1) return 1;    else return 0;}double mid,r,l,maxval;int main(){    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&n,&m);        maxval = 0;        for(int i = 0;i < n; i++)        {            scanf("%lf",&val[i]);            val[i] = val[i] * val[i] * PI;            maxval += val[i];        }        mid = maxval / (m + 1);        r = mid;        l = 0.0;        while(r - l > 1e-7)        {          if(check(mid))          {              l = mid;          }          else          {              r = mid;          }          mid=(l + r) / 2.0;        }        printf("%.4lf\n",mid);    }    return 0;}

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