pairs
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4157 Accepted Submission(s): 1481
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
2
5 5
-100
0
100
101
102
5 300
-100
0
100
101
102
Sample Output
3
10
题意
有x个数，求出这x个数中有多少对数相减的绝对值小于等于k
思路
AC代码
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <math.h>
#include <limits.h>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#include <set>
#include <string>
#define ll long long
#define ms(a) memset(a,0,sizeof(a))
#define pi acos(-1.0)
#define INF 0x3f3f3f3f
const double E=exp(1);
const int maxn=1e6+10;
using namespace std;
int a[maxn];
int main(int argc, char const *argv[])
{
ios::sync_with_stdio(false);
int t;
int n,k;
cin>>t;
while(t--)
{
cin>>n>>k;
for (int i = 1; i <=n; ++i)
{
cin>>a[i];
}
sort(a+1,a+n+1);
ll ans=0;
for(int i=1;i<=n;i++)
{
int l=lower_bound(a+1+i,a+1+n,a[i]-k)-a-1;
int r=upper_bound(a+i+1,a+1+n,a[i]+k)-a-1;
ans+=r-l;
}
cout<<ans<<endl;
}
return 0;
}