codeforces 1305C（鸽巢原理+同余）

 

 

 

 

 题意：给定n个数，n个数两两做差取绝对值，所有差的绝对值做累乘，再对m取模，求这个模的值。

思路：n=2e5，m=1000.如果n大于m，n个数分别%m肯定会出现两个数%n的余数相等，则他俩%m的余数相减肯定为0，则答案为0.n小于等于m时，直接暴力即可1e6

#include <iostream>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <string>
#include <map>
#include <iomanip>
#include <algorithm>
#include <queue>
#include <stack>
#include <set>
#include <vector>
//const int maxn = 1e5+5;
#define ll long long
#define inf  0x3f3f3f3f
#define FOR(i,a,b) for( int i = a;i <= b;++i)
#define bug cout<<"--------------"<<endl
#define P pair<int, int>
#define fi first
#define se second
 
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}

using namespace std;
const int maxn = 2e5 + 10;
int a[maxn];
int main()
{
    int n,m;
    ll ans = 1;
    cin>>n>>m;
    for(int i = 1;i <= n; ++i)
        cin>>a[i];
    if(n > m){
        cout<<0<<endl;
        return 0;
    }
    for(int i = 1;i <= n-1; ++i)
    {
        for(int j = i+1;j <= n; ++j)
        {
            ans = ((ans%m) *abs(a[i]-a[j])%m)%m;
        }
    }
    cout<<ans<<endl;
}