模数循环节

https://codeforces.com/contest/1305/problem/C

题意：给出n个数ai , 模数m 。 计算 ∏_1<=i<=j<=n|a[i] - a[j]|mod m

2 <= n <= 2e5 , 1 <= m <= 1000.

解法：直接暴力n方必超时。分析发现当n>m时，根据循环节性质必有两个数(a[i] - a[j])%mod = 0 ， 所以特判一下，时间复杂度

就为m方。

//#include<bits/stdc++.h>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <string>
#include <stdio.h>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <string.h>
#include <vector>
typedef long long ll ;
#define int ll
#define mod 1000000007
#define gcd __gcd
#define rep(i , j , n) for(int i = j ; i <= n ; i++)
#define red(i , n , j)  for(int i = n ; i >= j ; i--)
#define ME(x , y) memset(x , y , sizeof(x))
//ll lcm(ll a , ll b){return a*b/gcd(a,b);}
ll quickpow(ll a , ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;b>>=1,a=a*a%mod;}return ans;}
//int euler1(int x){int ans=x;for(int i=2;i*i<=x;i++)if(x%i==0){ans-=ans/i;while(x%i==0)x/=i;}if(x>1)ans-=ans/x;return ans;}
//const int N = 1e7+9; int vis[n],prime[n],phi[N];int euler2(int n){ME(vis,true);int len=1;rep(i,2,n){if(vis[i]){prime[len++]=i,phi[i]=i-1;}for(int j=1;j<len&&prime[j]*i<=n;j++){vis[i*prime[j]]=0;if(i%prime[j]==0){phi[i*prime[j]]=phi[i]*prime[j];break;}else{phi[i*prime[j]]=phi[i]*phi[prime[j]];}}}return len}
#define SC scanf
#define INF  0x3f3f3f3f
#define PI acos(-1)
#define pii pair<int,int>
#define fi first
#define se second
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
using namespace std;
const int N = 1e6+100;
const int maxn = 2e5+9;
int a[maxn];
void solve(){
    int n , m ;
    cin >> n >> m ;
    rep(i , 1 , n){
        cin >> a[i];
    }
    if(n > m){
        cout << 0 << endl;
        return ;
    }else{
        int ans = 1 ;
        rep(i , 1 , n){
            rep(j , i+1 , n){
                ans = ans % m * abs(a[i] - a[j]) % m ;
            }
        }
        cout << ans%m << endl;
    }
 
 
}
 
signed main()
{
    //ios::sync_with_stdio(false);
    //cin.tie(0); cout.tie(0);
    //int t ;
    //cin >> t ;
    //while(t--){
        solve();
    //}
}