Codeforce-Ozon Tech Challenge 2020-C. Kuroni and Impossible Calculation(鸽笼原理）

To become the king of Codeforces, Kuroni has to solve the following problem.

He is given n numbers a1,a2,…,an. Help Kuroni to calculate ∏1≤i<j≤n|ai−aj|. As result can be very big, output it modulo m.

If you are not familiar with short notation, ∏1≤i<j≤n|ai−aj| is equal to |a1−a2|⋅|a1−a3|⋅ … ⋅|a1−an|⋅|a2−a3|⋅|a2−a4|⋅ … ⋅|a2−an|⋅ … ⋅|an−1−an|. In other words, this is the product of |ai−aj| for all 1≤i<j≤n.

Input
The first line contains two integers n, m (2≤n≤2⋅105, 1≤m≤1000) — number of numbers and modulo.

The second line contains n integers a1,a2,…,an (0≤ai≤109).

Output
Output the single number — ∏1≤i<j≤n|ai−aj|modm.

Examples
inputCopy
2 10
8 5
outputCopy
3
inputCopy
3 12
1 4 5
outputCopy
0
inputCopy
3 7
1 4 9
outputCopy
1
Note
In the first sample, |8−5|=3≡3mod10.

In the second sample, |1−4|⋅|1−5|⋅|4−5|=3⋅4⋅1=12≡0mod12.

In the third sample, |1−4|⋅|1−9|⋅|4−9|=3⋅8⋅5=120≡1mod7.

这就是个鸽笼原理，m<=1000

#include <bits/stdc++.h>
using namespace std;
template <typename t>
void read(t &x)
{char ch = getchar();x = 0;t f = 1;while (ch < '0' || ch > '9')f = (ch == '-' ? -1 : f), ch = getchar();while (ch >= '0' && ch <= '9')x = x * 10 + ch - '0', ch = getchar();x *= f;
}#define wi(n) printf("%d ", n)
#define wl(n) printf("%lld ", n)
#define rep(m, n, i) for (int i = m; i < n; ++i)
#define rrep(m, n, i) for (int i = m; i > n; --i)
#define P puts(" ")
typedef long long ll;
#define MOD 1000000007
#define mp(a, b) make_pair(a, b)
#define N 200005
#define fil(a, n) rep(0, n, i) read(a[i])
//---------------https://lunatic.blog.csdn.net/-------------------//
int a;
int b[1005], flag;
vector<pair<int, int>> c;
int main()
{int n, m;read(n), read(m);for (int i = 0; i < n; i++){read(a);int mo = a % m;b[mo]++;c.push_back(make_pair(mo, a));if (b[mo] >= 2)flag = 1;// cout << mo << endl;}if (flag){puts("0");return 0;}int ans = 1;for (int i = 0; i < c.size(); i++){for (int j = i + 1; j < c.size(); j++){if (c[i].second > c[j].second){ans *= (c[i].first - c[j].first + m);}else{ans *= (-c[i].first + c[j].first + m);}ans %= m;}}cout << ans << endl;
}

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