P1102 A-B 数对

对数组升序排序，枚举A
(A-B=C)问题转化为对于A在A之前找等于(A-C)的值的个数，可以用二分做
复杂度：(O(nlogn))

#include<iostream>
#include<algorithm>

using namespace std;

#define LL long long

const int N = 200010;

int a[N];
int n, c;

int main(){
    cin >> n >> c;
    
    for(int i = 1; i <= n; i ++) cin >> a[i];
    
    sort(a + 1, a + n + 1);
    
    LL res = 0;
    for(int i = 2; i <= n; i ++){
        int l = 1, r = i - 1, x = a[i] - c;
        while(l < r){
            int mid = l + r >> 1;
            if(a[mid] >= x) r = mid;
            else l = mid + 1;
        }
        if(a[l] == x){
            int p = l;
            l = 1, r = i - 1;
            while(l < r){
                int mid = l + r + 1 >> 1;
                if(a[mid] <= x) l = mid;
                else r = mid - 1;
            }
            if(a[l] == x) res += l - p + 1;
        }
    }
    
    cout << res;
    
    return 0;
}