Codeforces 1305C Kuroni and Impossible Calculation

题目链接

题意，给你一个数组，求\(\prod_{1\le i<j\le n} |a_i - a_j|\)对\(m\)取模

分析数据，n是2e5, 而m是1000, 那肯定要从m下手，若\(m\le m\), 那么可以直接暴力求解, 若\(n>m\), 一定存在\(a_i, a_j\), 使得\(|a_i-a_j|\)一定能被\(m\)整除

#include<bits/stdc++.h>
using namespace std;
#define ms(x,y) memset(x, y, sizeof(x))
#define lowbit(x) ((x)&(-x))
typedef long long LL;
typedef pair<int,int> pii;

const int maxn = 1005;

LL buf[maxn];


void run_case() {
    int n, m;
    cin >> n >> m;
    if(m < n) {
        cout << "0\n";
        return;
    } else {
        for(int i = 1; i <= n; ++i) cin >> buf[i];
        LL ans = 1;
        for(int i = 1; i <= n; ++i)
            for(int j = i+1; j <= n; ++j) {
                ans = (ans * abs(buf[i] - buf[j])) % m;
            }
        cout << ans%m;
    }
}

int main() {
    ios::sync_with_stdio(false), cin.tie(0);
    cout.flags(ios::fixed);cout.precision(2);
    //int t; cin >> t;
    //while(t--)
    run_case();
    cout.flush();
    return 0;
}