BestCoder_Pairs

pairs

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 0    Accepted Submission(s): 0

Problem Description

John has n points on the X axis, and their coordinates are (x[i],0),(i=0,1,2,…,n−1). He wants to know how many pairs<a,b> that |x[b]−x[a]|≤k.(a<b)

 

Input

The first line contains a single integer T (about 5), indicating the number of cases.
Each test case begins with two integers n,k(1≤n≤100000,1≤k≤109).
Next n lines contain an integer x[i](−109≤x[i]≤109), means the X coordinates.

 

Output

For each case, output an integer means how many pairs<a,b> that |x[b]−x[a]|≤k.

 

Sample Input

25 5-10001001011025 300-1000100101102

 

Sample Output

310
简单水题!相信也不需要多说,二重循环一下就行.只是注意它每次减的数，是|x[b]−x[a]|,是两个数的绝对值.不能减多了也不能减少了！中文在下面......
<span style="font-family: Menlo, Monaco, Consolas, 'Courier New', monospace; font-size: 14px; line-height: 1.42857143; white-space: pre-wrap;">import java.io.*;import java.util.*;public class Main{public static void main(String[] args){// TODO Auto-generated method stubScanner input = new Scanner(System.in);int t = input.nextInt();for (int i = 0; i < t; i++){int n = input.nextInt();int a[] = new int[n];int sum = 0, flag = 0;int k = input.nextInt();for (int j = 0; j < n; j++){a[j] = input.nextInt();}for (int j = 0; j < n; j++){for (int l = j + 1; l < n; l++){if (a[j] - a[l] < 0){flag = -(a[j] - a[l]);} else{flag = a[j] - a[l];}if (flag <= k){sum++;}}}System.out.println(sum);}}}</span>

pairs

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 483    Accepted Submission(s): 234

问题描述

John 在X轴上拥有n个点，他们的坐标分别为(x[i],0),(i=0,1,2,…,n−1)。 他想知道有多少对<a,b>满足|x[b]−x[a]|≤k(a<b)。

输入描述

第一行包含一个正整数T(大约5)，表示有多少组数据。对于每一组数据，先读入两个数n,k(1≤n≤100000,1≤k≤109)。接下来n行，分别输入x[i](−109≤x[i]≤109)。

输出描述

对于每组数据，输出一行表示有多少对<a,b>满足|x[b]−x[a]|≤k。

输入样例

25 5-10001001011025 300-1000100101102

输出样例

310

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