POJ 1064 Cable master (二分搜索)

Cable master

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 25443		Accepted: 5471

Description

Inhabitants of the Wonderland have decided to hold a regional programming contest. The Judging Committee has volunteered and has promised to organize the most honest contest ever. It was decided to connect computers for the contestants using a "star" topology - i.e. connect them all to a single central hub. To organize a truly honest contest, the Head of the Judging Committee has decreed to place all contestants evenly around the hub on an equal distance from it. 
To buy network cables, the Judging Committee has contacted a local network solutions provider with a request to sell for them a specified number of cables with equal lengths. The Judging Committee wants the cables to be as long as possible to sit contestants as far from each other as possible. 
The Cable Master of the company was assigned to the task. He knows the length of each cable in the stock up to a centimeter,and he can cut them with a centimeter precision being told the length of the pieces he must cut. However, this time, the length is not known and the Cable Master is completely puzzled. 
You are to help the Cable Master, by writing a program that will determine the maximal possible length of a cable piece that can be cut from the cables in the stock, to get the specified number of pieces.

Input

The first line of the input file contains two integer numb ers N and K, separated by a space. N (1 = N = 10000) is the number of cables in the stock, and K (1 = K = 10000) is the number of requested pieces. The first line is followed by N lines with one number per line, that specify the length of each cable in the stock in meters. All cables are at least 1 meter and at most 100 kilometers in length. All lengths in the input file are written with a centimeter precision, with exactly two digits after a decimal point.

Output

Write to the output file the maximal length (in meters) of the pieces that Cable Master may cut from the cables in the stock to get the requested number of pieces. The number must be written with a centimeter precision, with exactly two digits after a decimal point. 
If it is not possible to cut the requested number of pieces each one being at least one centimeter long, then the output file must contain the single number "0.00" (without quotes).

Sample Input

4 11
8.02
7.43
4.57
5.39

Sample Output

2.00

题目大意：一共N条线段，如果截取出k条线段来，那么最长的长度有多长（线段可以不用完，线段只能截，不能接）。

此题用二分搜索搞，二分搜索详细分析见POJ 2456 Aggressive cows 二分搜索

那就来看一下，二分搜索里面的判断C( ) 函数怎么写：如果要截取长度为x的线段，那么长度为l的线段能截取出长度为x的线段的条数为：floor(l/x) 。只要对每条线段求出能截取的条数加起来的总和大于等于k则说明长度x可以满足。

然后就下来就能二分搜索了。

注意处理0.00，题目要求如果最终求的长度不能满足大于等于0.01则输入0.00.

对精度的处理看代码。

本题ac代码:

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>
using namespace std;
typedef __int64 ll;
#define INF 1e8
#define eps 1e-9

ll n,k;
double a[10005];

int C(double x)
{
	ll i,res=0;
	for(i=0;i<n;i++)
	{
		res+=(ll)(a[i]/x);
	}
	return res>=k;
}

int main()
{
	ll i,j,sum;
	double l,r,m;
	while(scanf("%I64d%I64d",&n,&k)!=EOF)
	{
		if(n==0 && k==0)
			break;
		sum=0;
		for(i=0;i<n;i++)
		{
			scanf("%lf",&a[i]);
			sum+=(int)floor(a[i]);
		}
		l=0;r=INF;
		for(i=0;i<100;i++)
		{
			m=(l+r)/2;
		//	printf("m=%.1f\n",m);
			if(C(m))
				l=m;
			else
				r=m;
		}
		if(fabs(l-0.01)<eps)
			printf("0.00\n");
		else
		printf("%.2f\n",floor(r*100)/100);
	}
	return 0;
}

一个for循环，循环100次，精度会精确到10^-30，完全可以满足题目的要求了。