HDU5178：pairs（二分法 & 尺取法）

pairs
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2497    Accepted Submission(s): 930

Problem Description
John has  points on the X axis, and their coordinates are . He wants to know how many pairs that 
 

Input
The first line contains a single integer  (about 5), indicating the number of cases.
Each test case begins with two integers .
Next  lines contain an integer , means the X coordinates.
 

Output
For each case, output an integer means how many pairs that .
 

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

Sample Output
3 10
 

Source
BestCoder Round #31
 

题意：

n个数a[i]，问有多少对的差（绝对值）<=k

分析：

暴力枚举每一个数a[i]，二分寻找每一个a[i]+k的个数

最近在学习尺取法，说一下这题的尺取做法

尺取法：其实这是一个升序的查找过程，找到了，后面就不可能有了，这样就ok了

#include <iostream>
#include <cstdio>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <algorithm>
#include <vector>
#include <string>
#include <cstring>
#include <sstream>
#define INF 100000000
using namespace std;
typedef long long ll;
const int N=100005;
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+n+1);
        ll ans=0;
        for(int i=1;i<=n;i++)
        {
            int pos=upper_bound(a+1+i,a+1+n,a[i]+k)-(a);
            ans+=pos-i-1;
        }
        printf("%lld\n",ans);
    }
    return 0;
}

# include <stdio.h>
# include <algorithm>
using namespace std;
 
int a[100001];
int main()
{
    int n, k, t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&k);
        long long ans = 0;
        for(int i=0; i<n; ++i)
            scanf("%d",&a[i]);
        sort(a, a+n);
        for(int l=0,r=1; l<n; ++l) 
        {
            while(r<n && a[r]-a[l]<=k) ++r;
            ans += r-l-1;
        }
        printf("%lld\n",ans);
    }
    return 0;
}