Y. The last 2 digits (#)

#include<bits/stdc++.h>
using namespace std;

int main(){
    int a,b,c,d,p;
    cin>>a>>b>>c>>d;
    a=a%100;    //why we are doing this is becuase the multipication might lead to overflow so we are here reducing their value according to contraints
    b=b%100;
    c=c%100;
    d=d%100;  

    p=a*b*c*d;
    
    //Now printing last two digits of p
    if(p%100<=9) cout<<0;       //you can also do this if(p%100<=9) cout<<"0"<<p;  else cout<<p%100;  Note : "0" is in string otherwise if taken in no c++/c will neglec it
    cout<<p%100;
    
    return 0;

}


LOGIC:
5854 X 5896 = 34515184
54 X 96 = 5184
Conclusion on taking a%100, the last two digits of product still remains the same


link - https://codeforces.com/group/MWSDmqGsZm/contest/219158/problem/Y