[hihocoder][Offer收割]编程练习赛45

互补二元组

Xi + Xj = Yi + Yj等价于Xi - Yi + Xj - Yj = 0 ，对每个二元组计算其x与y的差，每次加上其相反数的个数。

#pragma comment(linker, "/STACK:102400000,102400000")
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<vector>
#include<algorithm>
#include<iostream>
#include<map>
#include<queue>
#include<stack>
#include<string>
#include<functional>
#include<math.h>
//#include<bits/stdc++.h>
using namespace std;
typedef long long lint;
typedef vector<int> VI;
typedef pair<int, int> PII;
typedef queue<int> QI;


void makedata() {
    freopen("input.txt", "w", stdout);
    fclose(stdout);
}

map<int, int> mp;

int main() {
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
#endif
    //makedata();
    std::ios::sync_with_stdio(0), cin.tie(0);
    int n, x, y;
    lint ans = 0;
    cin >> n;
    for (int i = 0; i < n; i++) {
        cin >> x >> y;
        if (mp.find(y - x) != mp.end()) ans += mp[y - x];
        mp[x - y]++;
    }
    cout << ans << endl;
    return 0;
}

View Code