Aggressive cows (二分)
来源:互联网 发布:大学生论文单片机课题 编辑:程序博客网 时间:2024/06/05 05:01
Problem Description
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?
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 <br> <br>* 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 312849
Sample Output
3
题目大概:
求m只动物间的最大距离。
代码:
#include <iostream>#include <algorithm>#include <cmath>#include <cstdio>using namespace std;int n,k;int a[100005];int go(int m){ int h=a[1],sum=1; for(int i=2;i<=n;i++) { if(a[i]-h>=m){sum++; h=a[i];} } if(sum>=k)return 1; else return 0;}int main(){ scanf("%d%d",&n,&k); for(int i=1;i<=n;i++) { scanf("%d",&a[i]); } sort(a+1,a+n+1); int l=0,r=(a[n]-a[1])/(k-1); int mid; while(l<=r) { mid=(l+r)/2; if(go(mid)) { l=mid+1; } else { r=mid-1; } } printf("%d\n",l-1); return 0;}
阅读全文
0 0
- Aggressive cows(二分)
- Aggressive cows--(二分)
- Aggressive cows (二分)
- Aggressive cows (二分)
- POJ2456(Aggressive cows)(贪心+二分)
- POJ 2456 Aggressive cows(二分)
- POJ 2456 - Aggressive cows(二分)
- poj 2456 Aggressive cows(贪心+二分)
- POJ2456 Aggressive cows(二分搜索)
- poj 2456 Aggressive cows(贪心+二分)
- POJ 2456 Aggressive cows(二分)
- POJ 2456 Aggressive cows (二分搜索)
- POJ 2456 Aggressive cows(二分)
- POJ - 2456 Aggressive cows(二分查找)
- POJ2456 Aggressive cows 最大值最小化(二分)
- POJ 2456 Aggressive cows (贪心 + 二分)
- poj 2456 Aggressive cows (二分查找)
- POJ 2456 Aggressive cows (二分、贪心)
- LinkedList源码分析
- 【xml-ass】B站弹幕相关
- hibernate组件属性(当组件属性为集合的时候)
- 三:计算机进行小数运算时出错的原因
- javascript写各种排序算法
- Aggressive cows (二分)
- 使用Java反射机制访问类的私有属性
- angularJS中$apply()方法详解
- c#图像处理-图像预览全解
- 文章-自学的程序员如何找到好工作?
- hdu6103(尺取)
- 8.11 Closest 2574
- linux设备驱动之platform总线驱动学习
- Android TabLayout && 个性指示器