Ozon Tech Challenge 2020 (Div.1 + Div.2）C. Kuroni and Impossible Calculation

C. Kuroni and Impossible Calculation
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
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.

題意：
求出兩兩差的絕對值的乘積。

思路:

1. 第一眼2e5的數據認爲暴力不可行，以爲要推式子，後面想到了鴿巢原理。
2. 因爲m<=1000，所以當n>1000時，答案必定是0。
3. 爲什麼？因爲n>mod，必定出現a%mod=b%mod -> |a-b|%mod=0
4. 只要一個部分出現0，那麼乘積必定爲0 。
5. 所以n>1000的時候直接輸出0，其餘的O(n^2)也可以解決。

代碼:

#include<bits/stdc++.h>
#define ll long long
#define inf 0x3f3f3f3f
#define mod 1000000007
#define PI acos(-1)
#define fi first
#define se second
#define lowbit(x) (x&(-x))
#define mp make_pair
#define pb push_back
#define si size()
#define E exp(1.0)
#define fixed cout.setf(ios::fixed)
#define fixeds(x) setprecision(x)
using namespace std;
inline ll read(){ll s=0,w=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')w=-1;ch=getchar();}
while(ch>='0'&&ch<='9') s=s*10+ch-'0',ch=getchar();
return s*w;}void put1(){ puts("Yes") ;}void put2(){ puts("No") ;}
void debug(){printf("T   A   T\n");}void put3(){ puts("-1"); }ll qpf(ll a, ll b, ll p)
{ll ret = 0;while(b){if(b & 1) ret = (ret + a) % p;a = (a + a) % p;b >>= 1;}
return ret % p ;}ll qp(ll a, ll n, ll p){ll ret = 1;while(n){if(n & 1) ret = qpf(ret, a, p);
a = qpf(a, a, p);n >>= 1;}return ret % p ;}//θ=acos(L/2R);
 
const int manx=2e5+5;
 
ll b[manx];
int main(){
    ll n=read(),m=read(),ans=1;
    for(int i=1;i<=n;++i) b[i]=read();
    if(n>1000){
        cout<<0<<endl;
        return 0;
    }
    for(int i=1;i<=n;++i)
        for(int j=i+1;j<=n;++j)
            ans=(ans*abs(b[i]-b[j]))%m;
    cout<<ans%m<<endl;
    return 0;
}