poj 1064 Cable master 【二分 + 精度控制】

Cable master

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 29552		Accepted: 6246

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
题意：给你N段电缆，你需要把这些电缆分成等长的K段，问你能够分的最大长度。要求长度最短为1厘米。 输出结果保留两位小数。

思路很好想，二分查找长度区间即可。
注意精度问题：
对于数据0.009，它的输出为0.00。
对于数据1.67876，它的输出为1.67。到这想必你就恍然大悟了。

处理有两种方法
1，判断长度是否大于0.01，小于的话就输出0.00；若大于则对right*100向下取整，然后除以100即可。

代码实现：

if(right < 0.01) 
        printf("0.00\n");
else
        printf("%.2lf\n", floor(right*100) / 100);

2，判断结果的小数点后第三位小数，若大于或等于5，直接减去0.005即可。

代码实现：

if((int)(right*1000)%10 >= 5)//小数点后第三位是否大于或等于5 控制精度 
    right -= 0.005;

AC代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#define MAXN 10000+10
#define eps 1e-10
#define DD double
using namespace std;
DD a[MAXN];
int N, K;
int can(DD mid)
{
    int cnt = 0;
    for(int i = 0; i < N; i++)
        cnt += (int)(a[i] / mid);
    return cnt >= K;
}
int main()
{
    scanf("%d%d", &N, &K);
    DD sum = 0; DD right = 0;
    for(int i = 0; i < N; i++)
        scanf("%lf", &a[i]), right = max(right, a[i]);
    DD left = 0, mid;
    while(right - left >= eps)
    {
        mid = (left + right) / 2;//每次查询的长度
        if(can(mid))
            left = mid;
        else
            right = mid;
    }
    if((int)(right*1000)%10 >= 5)//小数点后第三位是否大于或等于5 控制精度 
    right -= 0.005;
        printf("%.2lf\n", (double)right);//hdojDIY里面GUN C++编译器 或者pojG++编译器请用%f输出 
    return 0;
}