POJ 1064 Cable master （二分）

题目链接: 传送门

Cable master

Time Limit: 1000MS     Memory Limit: 65536K

题目描述

思路

#include<iostream>
#include<cstdio>
#include<cmath>
#define EPS 1e-6
const int INF = 100000;
using namespace std;
int N,K;
double ans[10005];

bool binary(double x)
{
int sum = 0;
for (int i = 0;i < N;i++)
{
sum += (int)(ans[i] / x);
}
return sum >= K;
}

int main()
{
while (scanf("%d%d",&N,&K) != EOF)
{
for (int i = 0;i < N;i++)
{
scanf("%lf",&ans[i]);
}

double left = 0,right = INF,mid;

while (right - left > EPS)
{
mid = (left + right)/2;
if (binary(mid))
{
left = mid;
}
else
{
right = mid;
}
}

/*if (mid < 1)
{
printf("0.00\n");
}
else
{
printf("%.2lf\n",mid);
}*/

printf("%.2f\n",floor(right*100)/100); //防止四舍五入
}

return 0;
}