C. Kuroni and Impossible Calculation

C. Kuroni and Impossible Calculation

1. The meaning of problems

   A given number of columns \ (A [] \) , and a modulus \ (m \) , calculated \ ({Π_. 1 <= I <J <= n-} \) \ (|a [I] -a [ j] | \)

   数据范围：\(1<=n<=2∗10^5,1<=m<=1000,0<=a[i]<=10^9\)

2. Thinking

   It is easy to think of (O (n ^ 2) \ ) \ practice, but to see the range of n, obviously time out.
   M ranges will be seen, is not a very large number of fixed mass as modulus, it should be a breakthrough m.
   The m, can be divided into two cases.

   1. \ (n <= m \) , n is such a range less than 1000, can be \ (O (n ^ 2) \) solution
   2. \ (n-> m \) , m has finished after molding up to m different modulus, and the counted number is greater than the number of columns m, there must be at least two numbers, \ (A [I] ≡a [J] MODM \) , then the \ (|a [I] -a [J] |≡0modm \) .
      the final product is 0.

3. Code

   #include<bits/stdc++.h>
   using namespace std;
   const int maxn = 2e5+10;
   typedef long long ll;
   ll a[maxn];
   int book[1010];
   int main(void){
    int n,m;
    scanf("%d%d",&n,&m);
    for(int i = 1; i <= n; i++){
        scanf("%d",&a[i]);
        if(book[a[i]%m]++){
            printf("0\n");
            return 0;
        }
    } 
    ll ans = 1;
    for(int i = 1; i <= n; i++){
        for(int j = i+1; j <= n; j++){
            ans *= abs(a[i]-a[j]);
            ans %= m; 
            if(ans==0){
                printf("%lld\n",ans);
                return 0;
            }
        }
    }
    printf("%lld\n",ans);
    return 0;
   }