Cable master 求电缆的最大长度（二分法)
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.
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 input consists of several testcases. The first line of each testcase contains two integer numbers 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 centimeter and at most 100 kilometers in length. All lengths in the input are written with a centimeter precision, with exactly two digits after a decimal point.
The input is ended by line containing two 0's.
The input is ended by line containing two 0's.
Output
For each testcase write to the output 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 must contain the single number "0.00" (without quotes).
If it is not possible to cut the requested number of pieces each one being at least one centimeter long, then the output must contain the single number "0.00" (without quotes).
Sample Input
4 11
8.02
7.43
4.57
5.39
0 0
Sample Output
2.00
题意：有N条绳子，他们的长度分别是Li。如果从他们中切割出K条长度相同的绳子的话，这K条绳子每条最长能有多长？答案保留2位小数。（例题）
题解
也就是求一个x , l1/ x +l2/x +l3/x +.....=K，求最大的x。求的过程中中间值x ，如果>=k也时 ，要求最大的x.
条件C（x）=可以得到K条长度为x的绳子
区间l=0,r等于无穷大，二分，判断是否符合c(x) C(x)=(floor(Li/x)的总和大于或等于K
题解
也就是求一个x , l1/ x +l2/x +l3/x +.....=K，求最大的x。求的过程中中间值x ，如果>=k也时 ，要求最大的x.
条件C（x）=可以得到K条长度为x的绳子
区间l=0,r等于无穷大，二分，判断是否符合c(x) C(x)=(floor(Li/x)的总和大于或等于K
输入电缆的个数n，和要分的份数k，求出所有电缆的长度和sum，在区间[0,sum/k]内用二分法求出最大的长度。
#include<iostream> #include<iomanip> #include<cstdio> #include<math.h> using namespace std; #define st 1e-8//10^-8 double a[10010]; int k, n; int f(double x) { int cnt = 0; for (int i = 0; i < n; i++) { cnt += (int)(a[i] / x); //括号不能少 } return cnt; } int main() { double sum; scanf("%d%d", &n, &k); sum = 0; for (int i = 0; i < n; i++) { scanf("%lf", &a[i]); sum += a[i]; } sum = sum / k;//平均值一定会比最终取得长度大 double l = 0, r = sum, mid; while (fabs(l - r)>st)//控制精度 { mid = (l + r) / 2; if (f(mid) >= k)//不断逼近，找到可以满足切割数量下的最大长度 l = mid; else r = mid; } r=r*100; printf("%.2f ", floor(r)/100);//向下取整 } // 4 2542 // 8.02 // 7.43 // 4.57 // 5.39 // 0.00