Cable master
成绩 | 10 | 开启时间 | 2018年02月11日 星期日 15:00 |
折扣 | 0.8 | 折扣时间 | 2018年12月30日 星期日 18:00 |
允许迟交 | 否 | 关闭时间 | 2018年12月30日 星期日 20:00 |
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 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).
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
Source
Northeastern Europe 2001
二分答案。开始l=0,r=(所有之和)/k
因为精度的问题,直接选择了二分100次。
输出的时候加上floor函数,向下取整
#include<cstdio>
#include<algorithm>
#include<string>
#include<cmath>
using namespace std;
#define INF 100005
#define exp 1e-5
#define maxn 10005
double a[maxn]; //存电缆的长度
int n, need;
int check(double x)
{
int unit = 0;
for (int i = 0; i < n; i++)
unit += (int)(a[i] / x);
if (unit >= need)
return 1;
else
return 0;
}
int main()
{
freopen("1.txt", "r", stdin);
while (scanf("%d %d", &n, &need) != EOF)
{
if (n == 0 && need == 0)
return 0;
double sum = 0;
for (int i = 0; i < n; i++)
{
scanf("%lf", &a[i]);
sum += a[i];
}
double lb = 0, ub = INF;
for (int i = 0; i<100; i++) //限定次数吧,别eps了,orz
{
double mid = (lb + ub) / 2;
if (check(mid))
lb = mid;
else
ub = mid;
}
printf("%.2f\n", floor(ub * 100) / 100);
}
return 0;
}