Cable master
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 35269 | Accepted: 7513 |
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
题目大意：有n根电缆，你需要k条等同长度的电缆，最后得到的电缆长度最长是多少。
思路分析：初学二分，做这道题的时候感觉和之前做的一道题非常相似，以为可以轻松切掉，可是在做的时候还是出现了问题，
正确的思路应该是化为厘米，然后用整数二分，如果直接用小数二分最后会出现问题四舍五入，对于这些数据
4 2540
8.02
7.43
4.57
5.39 =>0.01 4 2542
8.02
7.43
4.57
5.39 =>0.00
化为整数进行处理则可以避免这些问题，另外要注意二分上下限，下限自然是0，而上限你可以用sum/k,也可以用a[n-1]，当出现sum的时候，
就会超过int数据范围，要用__int64,如果用a[n-1]为上界就不需要开__int64了，再就是写函数时要判断是否有死循环，我就写错了，狂
TLE不止orz.
代码：
#include<map>
#include<queue>
#include<stack>
#include<cmath>
#include<cstdio>
#include<vector>
#include<string>
#include<cstdlib>
#include<cstring>
#include<climits>
#include<iostream>
#include<algorithm>
#include <cmath>
#define LL long long
using namespace std;
const int maxn=10000+100;
const double pi=acos(-1.0);
double a[maxn];
int b[maxn];
int n,k;
__int64 sum;
bool check(int x)
{
int t=0;
for(int i=0;i<n;i++)
{
int m=b[i];
while(m>=x)
{
m-=x;
t++;
if(t>=k) return true;
}
}
return false;
}
int main()
{
while(scanf("%d%d",&n,&k)!=EOF)
{
sum=0;
for(int i=0;i<n;i++)
{
scanf("%lf",&a[i]);
b[i]=a[i]*100;//将单位化为整数厘米
sum+=b[i];
}
sort(b,b+n);
int l=0,r=b[n-1];
int ans=0;
while(l<=r)
{
int mid=(l+r)/2;
if(check(mid)) ans=mid,l=mid+1;
else r=mid-1;
}
printf("%.2lf\n",ans*0.01);
}
return 0;
}