同余，余就完了

#include<iostream>
#include<cmath>
using namespace std;

long long int n,c,ans=1,a[119];

int main(){
	while(cin>>n>>c){
		ans=1;
		for(int i=0;i<n;i++){
			cin>>a[i];
			ans*=(a[i]%c);
			ans%=c;
		}
		 cout<<ans<<endl;
	}
	 return 0;
}

只需要每输入一个数，将它按照相关步骤走就对了：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e6+5;
int main()
{
	ll n,m;
	while(cin>>n>>m)
	{
		ll pp=1;
		while(n--)
		{
			ll a;
			cin>>a;
			pp=((a%m)*(pp%m))%m;
		}	
	cout<<pp<<endl;

	}
	return 0;
}

#include<stdio.h>
int main()
{
    long long n,c,ans = 1;;
    int count;
	while(~scanf("%lld%lld",&n,&c))
	{
		long long a[117];
		for(int i = 0;i < n;i ++)
		{
			scanf("%lld",&a[i]);
		}
		ans = ((a[0]%c) * (a[1]%c))%c;
		for(int i = 2;i < n; i++)
		{	
		    a[i] %= c;
			ans *= a[i]; 
			ans %= c;
			a[i] = 0;
		}
		printf("%lld\n",ans);
	}
	return 0;
 } 

#include<bits/stdc++.h>
using namespace std;
#define ll long long
ll a[205];
int main(){
	ll n,c,ans;
	while(scanf("%lld%lld",&n,&c)!=EOF){
		for(int i=0;i<n;i++){
				cin>>a[i];
			}
		for(int i=0;i<n;i++){
			a[i]=a[i]%c;
		}
		ans=((a[0]%c)*(a[1]%c))%c;
		for(int i=2;i<n;i++){
			ans=((ans%c)*(a[i]%c))%c;
		}
		cout<<ans<<endl;
	}
	return 0;
}