CodeForces - 1189E Count Pairs（平方差）

Count Pairs

You are given a prime number pp, nn integers a1,a2,…,ana1,a2,…,an, and an integer kk.

Find the number of pairs of indexes (i,j)(i,j) (1≤i<j≤n1≤i<j≤n) for which (ai+aj)(a2i+a2j)≡kmodp(ai+aj)(ai2+aj2)≡kmodp.

Input

The first line contains integers n,p,kn,p,k (2≤n≤3⋅1052≤n≤3⋅105, 2≤p≤1092≤p≤109, 0≤k≤p−10≤k≤p−1). ppis guaranteed to be prime.

The second line contains nn integers a1,a2,…,ana1,a2,…,an (0≤ai≤p−10≤ai≤p−1). It is guaranteed that all elements are different.

Output

Output a single integer — answer to the problem.

Examples

Input

3 3 0
0 1 2

Output

1

Input

6 7 2
1 2 3 4 5 6

Output

3

Note

In the first example:

(0+1)(02+12)=1≡1mod3(0+1)(02+12)=1≡1mod3.

(0+2)(02+22)=8≡2mod3(0+2)(02+22)=8≡2mod3.

(1+2)(12+22)=15≡0mod3(1+2)(12+22)=15≡0mod3.

So only 11 pair satisfies the condition.

In the second example, there are 33 such pairs: (1,5)(1,5), (2,3)(2,3), (4,6)(4,6).

 

 

题意：

找出有几组数对满足i<j，且(ai+aj)(ai^2+aj^2)≡k mod p

思路：

如果两两找O(n^2)肯定超时，所以想到需要让i与j分离，即i与j分列等式两边，这样只需扫一遍。

两边同乘(ai-aj)，制造平方差，整理后得(ai^4-aj^4)==k(ai-aj)，即ai^4-k*ai==aj^4-k*aj

排序后找相同对即可。

 

#include<bits/stdc++.h>
#define MAX 300005
using namespace std;
typedef long long ll;
 
ll a[MAX];
 
int main()
{
    int t,m,i,j;
    ll n,p,k;
    scanf("%I64d%I64d%I64d",&n,&p,&k);
    for(i=1;i<=n;i++){
        scanf("%I64d",&a[i]);
        a[i]=(a[i]*a[i]%p*a[i]%p*a[i]%p-k*a[i]%p+p)%p;
    }
    sort(a+1,a+n+1);
    ll c=0,ans=0;
    for(i=2;i<=n;i++){
        if(a[i-1]==a[i]){
            c++;
            ans+=c;
        }
        else c=0;
    }
    printf("%I64d\n",ans);
    return 0;
}