Problem Description

John has nn 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><a,b> that

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

Sample Input

2
5 5
-100
0
100
101
102
5 300
-100
0
100
101
102

Sample Output

3
10

題意：給出 n 個數，問有多少對 <a,b> 滿足

思路：尺取法

將 去掉絕對值，得：

對 n 個數排序后使用，對相鄰的數尺取即可

Source Program

#include<iostream> #include<cstdio> #include<cstdlib> #include<string> #include<cstring> #include<cmath> #include<ctime> #include<algorithm> #include<utility> #include<stack> #include<queue> #include<vector> #include<set> #include<map> #define PI acos(-1.0) #define E 1e-9 #define INF 0x3f3f3f3f #define N 100001 #define LL long long const int MOD=20091226; const int dx[]= {-1,1,0,0}; const int dy[]= {0,0,-1,1}; using namespace std; LL a[N]; int main(){int t;scanf("%d",&t);while(t--){int n;LL k;scanf("%d%lld",&n,&k);for(int i=1;i<=n;i++)scanf("%lld",&a[i]);sort(a+1,a+1+n);LL res=0;int l=1,r=1;while(l<=n){while(r+1<=n&&a[r+1]-a[l]<=k)r++;res+=(r-l);l++;}printf("%lld\n",res);}return 0; }

?

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