zoukankan      html  css  js  c++  java
  • POJ

    Farmer John has built a new long barn, with N (2 <= N <= 100,000) stalls. The stalls are located along a straight line at positions x1,...,xN (0 <= xi <= 1,000,000,000). 

    His C (2 <= C <= N) cows don't like this barn layout and become aggressive towards each other once put into a stall. To prevent the cows from hurting each other, FJ want to assign the cows to the stalls, such that the minimum distance between any two of them is as large as possible. What is the largest minimum distance?

    Input

    * Line 1: Two space-separated integers: N and C 

    * Lines 2..N+1: Line i+1 contains an integer stall location, xi

    Output

    * Line 1: One integer: the largest minimum distance

    Sample Input

    5 3
    1
    2
    8
    4
    9

    Sample Output

    3

    Hint

    OUTPUT DETAILS: 

    FJ can put his 3 cows in the stalls at positions 1, 4 and 8, resulting in a minimum distance of 3. 

    Huge input data,scanf is recommended.
     1 #include <iostream>
     2 using namespace std;
     3 #include<string.h>
     4 #include<set>
     5 #include<stdio.h>
     6 #include<math.h>
     7 #include<queue>
     8 #include<map>
     9 #include<algorithm>
    10 #include<cstdio>
    11 #include<cmath>
    12 #include<cstring>
    13 #include <cstdio>
    14 #include <cstdlib>
    15 #include<stack>
    16 #include<vector>
    17 long long n,k;
    18 long long a[110000];
    19 long long panduan(long long zhi)
    20 {
    21     int add=1;
    22     int kaishi=1,jieshu=1;
    23     while(1)
    24     {
    25         if(a[jieshu]-a[kaishi]<zhi&&jieshu<=n)
    26         {
    27             jieshu++;
    28         }
    29         else if(a[jieshu]-a[kaishi]>=zhi)
    30         {
    31             add++;
    32             kaishi=jieshu;
    33         }
    34         else
    35             break;
    36     }
    37     //cout<<zhi<<"_"<<add<<endl;
    38     return add;
    39 }
    40 int  main()
    41 {
    42     scanf("%lld %lld",&n,&k);
    43     for(long long  i=1;i<=n;i++)
    44         scanf("%lld",&a[i]);
    45     sort(a+1,a+1+n);
    46     long long kaishi=0,jieshu=1e18;
    47     long long mid;
    48     long long t;
    49     while(kaishi<jieshu-1)
    50     {
    51         //cout<<kaishi<<"_"<<jieshu<<endl;
    52         mid=(kaishi+jieshu)>>1;
    53         if(panduan(mid)>=k)
    54             kaishi=mid;
    55         else
    56             jieshu=mid;
    57     }
    58     cout<<kaishi<<endl;
    59     return 0;
    60 }
    View Code
     
  • 相关阅读:
    Android四大组件--事务详解(转)
    Android课程---关于数据存储的学习(3)之数据库和事务
    初学DW资料——js的prompt的返回值
    初学DW资料——target=的理解
    初学JAVA资料——链表
    初学JAVA资料——哈希表
    初学JAVA资料——线程
    初学JAVA——代码练习(验证字符串结束字符)
    初学JAVA——代码练习(数学运算)
    初学JAVA——代码练习(字符串)
  • 原文地址:https://www.cnblogs.com/dulute/p/7966712.html
Copyright © 2011-2022 走看看