Time Limit: 1000MS | ? | Memory Limit: 10000K |
Total Submissions: 21071 | ? | Accepted: 4542 |
Description
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
Output
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
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解題思路:
N條繩子,長度為別為li,要截成長度相等的K段,問切成小段的最大長度是多少。有的繩子能夠不切。
也就是求一個x ,?? l1/ x +l2/x +l3/x +.....=K,求最大的x。
求的過程中中間值x 。假設>k也是符合題意的。要求最大的x。==k.
條件C(x)=能夠得到K條長度為x的繩子
區間l=0,r等于無窮大,二分。推斷是否符合c(x) C(x)=(floor(Li/x)的總和大于或等于K
代碼:
#include <iostream> #include <iomanip> #include <stdio.h> #include <cmath> using namespace std; const int maxn=10003; const int inf=0x7fffffff; double l[maxn]; int n,k; bool ok(double x)//推斷x是否可行 { int num=0; for(int i=0;i<n;i++) { num+=(int)(l[i]/x); } return num>=k;//被分成的段數大于等于K才可行 } int main() { cin>>n>>k; for(int i=0;i<n;i++) scanf("%lf",&l[i]); double l=0,r=inf; for(int i=0;i<100;i++)//二分,直到解的范圍足夠小 { double mid=(l+r)/2; if(ok(mid)) l=mid; else r=mid; } cout<<setiosflags(ios::fixed)<<setprecision(2)<<floor(l*100)/100;//l和r最后相等 return 0; }
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