A. Reconnaissance
2016-04-21 19:03
148 查看
A. Reconnaissancetime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputAccording to the regulations of Berland's army, a reconnaissance unit should consist of exactly two soldiers. Since these two soldiers shouldn't differ much, their heights can differ by at most d centimeters. Captain Bob has n soldiers in his detachment. Their heights area1,?a2,?...,?an centimeters. Some soldiers are of the same height. Bob wants to know, how many ways exist to form a reconnaissance unit of two soldiers from his detachment.Ways (1,?2) and (2,?1) should be regarded as different.InputThe first line contains two integers n and d (1?≤?n?≤?1000,?1?≤?d?≤?109) — amount of soldiers in Bob's detachment and the maximum allowed height difference respectively. The second line contains n space-separated integers — heights of all the soldiers in Bob's detachment. These numbers don't exceed 109.OutputOutput one number — amount of ways to form a reconnaissance unit of two soldiers, whose height difference doesn't exceed d.Sample test(s)input
5 10 10 20 50 60 65output
6input
5 1 55 30 29 31 55output
6w
/* *********************************************** Author : Created Time :2015/6/14 11:33:10 File Name :7.cpp ************************************************ */ #include <iostream> #include <cstring> #include <cstdlib> #include <stdio.h> #include <algorithm> #include <vector> #include <queue> #include <set> #include <map> #include <string> #include <math.h> #include <stdlib.h> #include <iomanip> #include <list> #include <deque> #include <stack> #define ull unsigned long long #define ll long long #define mod 90001 #define INF 1<<30 #define maxn 10000+10 #define cle(a) memset(a,0,sizeof(a)) const ull inf = 1LL << 61; const double eps=1e-5; using namespace std; int a[maxn]; int main() { #ifndef ONLINE_JUDGE //freopen("in.txt","r",stdin); #endif //freopen("out.txt","w",stdout); int n,d; while(cin>>n>>d){ for(int i=1;i<=n;i++) cin>>a[i]; int ans=0; for(int i=1;i<=n;i++){ for(int j=1;j<=n;j++){ if(abs(a[i]-a[j])<=d&&i!=j)ans++; } } cout<<ans<<endl; } return 0; }
相关文章推荐
- HDU 1839 Delay Constrained Maximum Capacity Path
- HDU1789 Doing Homework again(贪心)
- 爬楼梯的问题Climbing Stairs
- 十、责任链模式Chain of Responsibility(行为型模式)
- http://blog.csdn.net/holybin/article/details/46955185
- 服务器端主动关闭连接, 产生的TIME_WAIT状态为什么会占用服务端大量端口?
- HoloLens开发手记-硬件细节 Hardware Detail
- git修改提交的用户名和Email
- 03-树3 Tree Traversals Again
- 服务器TIME_WAIT和CLOSE_WAIT详解和解决办法
- 正确使用Block避免Cycle Retain和Crash
- lightoj 1076 - Get the Containers 【二分 + 贪心】
- Failed to read artifact descriptor进行安装maven项目出错
- AI编辑SVG格式的相关问题
- paip.自适应网页设计 跟 响应式 设计的区别跟原理and实践总结
- 图解RAID 0, RAID 1, RAID 5, RAID 10
- go gomail
- R语言中的正则表达式(转载:http://blog.csdn.net/duqi_yc/article/details/9817243)
- kaidi中 install cuda
- stl map用法和make_pair函数