C题 Kuroni and Impossible Calculation

题目

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

If you are not familiar with short notation, ∏1≤i

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

Examples
Input
2 10
8 5
Output
3
Input
3 12
1 4 5
Output
0
Input
3 7
1 4 9
Output
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.

题意

求解 1≤ i 1≤ i… |a1 - an|* |a2 - a3| * |a2 - a4|… |a2 - an|*… * |an-1 - an|. 言外之意，求解 |ai - aj| 乘积。(1 ≤ i < j ≤ n)

思路

根据雀巢定理可得，当n>m的时候总有两个数%m 相等，根据同余定理得，他俩的差%m等于0
m最大是1000，所以n>m 的时候或者n>1000的时候答案就是0
剩下的就1000 暴力写就完事

代码

#include 
#include 
typedef long long ll;
using namespace std;
const int maxn = 200000+100;
int a[maxn];
int main()
{
	ios::sync_with_stdio(0);
	int n,m;
	cin>>n>>m;
	for(int i=1;i<=n;i++) cin>>a[i];
	if(n>=m+1) cout<<"0\n";
	else
	{
		ll ans = 1;
		for(int i=1;i<=n-1;i++)
		for(int j=i+1;j<=n;j++)
			ans = ans * abs(a[i] - a[j]) % m;
		cout<<ans % m<<"\n";
	}
        return 0;
} 
/*
n>m的时候，根据抽屉定理，总会有两个数字mod m相等，因为如果前m个都没
重复的话， 第m+1个肯定重复了 
*/