codeforces round#478D Ghosts

Open the transmission gate

Ghosts live in harmony and peace, they travel the space without any purpose other than scare whoever stands in their way.

There are n ghosts in the universe, they move in the OXY plane, each one of them has its own velocity that does not change in time: $\vec{V} =\vec{V_{x_i}}+\vec{V_{y_j}}$ where $\vec{V_{x_i}}$ is its speed on the x-axis and $\vec{V_{y_j}}$ is on the y-axis.

A ghost $i$ has experience value $EX_i$ , which represent how many ghosts tried to scare him in his past. Two ghosts scare each other if they were in the same cartesian point at a moment of time.

As the ghosts move with constant speed, after some moment of time there will be no further scaring (what a relief!) and the experience of ghost kind $GX=\sum_{i=1}^{n}EX_i$ will never increase.

Tameem is a red giant, he took a picture of the cartesian plane at a certain moment of time T , and magically all the ghosts were aligned on a line of the form y=a⋅x+b. You have to compute what will be the experience index of the ghost kind $GX$ in the indefinite future, this is your task for today.

Note that when Tameem took the picture, $GX$ may already be greater than 0, because many ghosts may have scared one another at any moment between $[−∞,T]$ .

Input

The first line contains three integers n, a and b (1≤n≤200000, 1≤|a|≤109, 0≤|b|≤109) — the number of ghosts in the universe and the parameters of the straight line.Each of the next n lines contains three integers $x_i$ , $V_{x_i}$ , $V_{y_i}$ (−109≤xi≤109, −109≤Vxi,Vyi≤109), where $x_i$ is the current x-coordinate of the i-th ghost (and $y_i=a⋅x_i+b$ ).

It is guaranteed that no two ghosts share the same initial position, in other words, it is guaranteed that for all (i,j) xi≠xj for i≠j.

Output

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Output one line: experience index of the ghost kind GX in the indefinite future.

Examples
Input

4 1 1
1 -1 -1
2 1 1
3 1 1
4 -1 -1

Output

8

Input

3 1 0
-1 1 0
0 0 -1
1 -1 -2

Output

6

Input

3 1 0
0 0 0
1 0 0
2 0 0

Output

0

Note

There are four collisions (1,2,T−0.5), (1,3,T−1), (2,4,T+1), (3,4,T+0.5), where (u,v,t) means a collision happened between ghosts u and v at moment t. At each collision, each ghost gained one experience point, this means that GX=4⋅2=8.

In the second test, all points will collide when t=T+1.

这里写图片描述

The red arrow represents the 1-st ghost velocity, orange represents the 2-nd ghost velocity, and blue represents the 3-rd ghost velocity.

【translation】

给你一条直线，在某个时刻的时候所有的鬼都在这个直线下，每个鬼都有自己的速度，问你有多少对鬼相遇过。

【solution】

其实说白了就是推公式。给一个队友的证明：传送门

【code】

/*
队友是用的map，我用的是数组模拟，他写起来简单，我稍微快一点点
*/
#include <bits/stdc++.h>
using namespace std;
const int maxm = 2e5+10;
long long num[maxm];
const int mod = 1e9+7;
long long nouse[maxm];
int main() {
    //freopen("C:\\Users\\ACM2018\\Desktop\\in.txt","r",stdin);
    long long n,a;
    long long ans = 0;
    scanf("%lld%lld%*lld",&n,&a);
    for(int i = 0;i<n;i++){
        long long vx,vy;
        scanf("%*d%lld%lld",&vx,&vy);
        nouse[i] = vx*mod+vy;
        num[i] = -vy+a*vx;
    }
    sort(nouse,nouse+n);
    nouse[n] = 2e18;
    long long has = 1;
    long long now = nouse[0];
    for(int i = 1;i<=n;i++){
        if(nouse[i]==now) has++;
        else{
            ans-=has*(has-1);
            has = 1;
            now = nouse[i];
        }
    }
    sort(num,num+n);
    has = 1;
    now = num[0];
    num[n] = 2e18;
    for(int i = 1;i<=n;i++){
        if(num[i]==now){
            has++;
        }
        else{
            ans+=has*(has-1);
            has = 1;
            now = num[i];
        }
    }
    printf("%lld\n",ans);
    return 0;
}

codeforces round#478D Ghosts

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