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.

這就是個(gè)鴿籠原理，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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