Codeforces Round #478 (Div. 2) D - Ghosts

D. Ghosts

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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: $\overrightarrow{V}=V_x\overrightarrow{i}+V_y\overrightarrow{j}$ where $V_x$ is its speed on the $x$-axis and $V_y$ is on the $y$-axis.

A ghost $i$ has experience value $E{X}_i$, 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 $GX={\sum }_{i=1}^{n}E{X}_i$ 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 $y=a\cdot x+b$. 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 $[-\infty ,T]$.

Input

The first line contains three integers $n$, $a$ and $b$ ($1\le n\le 200000$, $1\le |a|\le {10}^9$, $0\le |b|\le {10}^9$) — the number of ghosts in the universe and the parameters of the straight line.

Each of the next $n$ lines contains three integers $x_i$, $V_{xi}$, $V_{yi}$ ($-{10}^9\le x_i\le {10}^9$, $-{10}^9\le V_{xi},V_{yi}\le {10}^9$), where $x_i$ is the current $x$-coordinate of the $i$-th ghost (and $y_i=a\cdot x_i+b$).

It is guaranteed that no two ghosts share the same initial position, in other words, it is guaranteed that for all $(i,j)$ $x_i\ne x_j$ for $i\ne j$.

Output

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

Examples

input

Copy

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

output

Copy

8

input

Copy

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

output

Copy

6

input

Copy

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

output

Copy

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 $GX=4\cdot 2=8$.

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

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

题意:有n个点具有恒定的初速度，设每个点和其他点相遇次数为sumi,求Σsumi(1<=i<=n).

题解:

#include <bits/stdc++.h>
#include <ext/hash_map>
#include <ext/hash_set>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/priority_queue.hpp>
using namespace std;
using namespace __gnu_cxx;
using namespace __gnu_pbds;
#define XINF INT_MAX
#define INF 0x3F3F3F3F
#define MP(X,Y) make_pair(X,Y)
#define PB(X) push_back(X)
#define REP(X,N) for(int X=0;X<N;X++)
#define REP2(X,L,R) for(int X=L;X<=R;X++)
#define DEP(X,R,L) for(int X=R;X>=L;X--)
#define CLR(A,X) memset(A,X,sizeof(A))
#define IT iterator
#define RIT reverse_iterator
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef vector<PII> VII;
typedef vector<int> VI;
typedef tree<PII, null_type, greater<PII>, rb_tree_tag, tree_order_statistics_node_update > rb_tree_set;
typedef tree<PII, PII, greater<PII>, rb_tree_tag, tree_order_statistics_node_update > rb_tree;
#define PQ std::priority_queue
#define HEAP __gnu_pbds::priority_queue
#define X first
#define Y second
#define lson(X) ((X)<<1)
#define rson(X) ((X)<<1|1)
#define EPS 10e-6
#define pi acos(-1)
#define N 300000
map<PII, int> mp;
map<ll , int> num;

int main()
{
    int n;
    ll a, b;
    scanf("%d%lld%lld", &n, &a, &b);
    int x, vx, vy;
    ll ans = 0;
    REP(i, n) {
        scanf("%d%d%d", &x, &vx, &vy);
        ans += num[a*vx-vy]-mp[MP(vx, vy)];
        num[a*vx-vy] ++;
        mp[MP(vx, vy)] ++;  //处理同向的点
    }
    printf("%lld\n", ans*2);
    return 0;
}