Wow! My very first CP blog

/* @ShifatJahanShifa - this problem is not from any online judge. - the idea of the problem is : - certain amount of matters (T-01) ( human, fruits, flowers anything can be) share a characteristic (T-02) ; characteristic be like:- days (30), weekday (7), type of matters like ( fruits, flowers anything can be); of numbered ranging (0 to 64); (if we express the bitmask as a number. otherwise we can use bitsets for large number of bits (>64)) - what i will de is: - for every element of (T-01), i will create an integer mask number which bits will be set corresponding to the position in (T-02); Example; (T-01) - {1,2,3,4,5} # human (T-02) - {1,2,3} # fruits type so, human[1]={1,2} human[2]={2} human[3]={2,3} human[4]={3} human[5]={1,2,3} then, generated mask for each human: human[1]= 0 1 1 =3 human[2]= 0 1 0 =2 human[3]= 1 1 0 =6 human[4]= 1 0 0 =4 human[5]= 1 1 1 =7 - suppose the question is: in which pair, there is maximum common fruits?? - after the above masking process, iterating over each pair, and opration between their masking value, counting the set bit of the resultant value; I can find the answer (*.*). input format: n-(number of T-01), m-(number of (T-02)) k-(total have fruits), f1...fk-(fruits type number) . . . n ***** test cases ****** 1. 5 3 2 1 2 1 2 2 2 3 1 3 3 1 2 3 output: 2 2. 7 5 3 2 3 5 2 4 5 3 1 2 3 3 1 4 5 1 3 4 1 3 4 5 3 1 3 5 output: 3 ** constraints in input is simple, suppose int value will be accepted. */ #include <bits/stdc++.h> using namespace std; int main() { int n,m,k,e,mask;cin>>n>>m; vector<int>maskNumber; for (int i = 0; i < n; i++) { cin>>k; mask=0; while (k--) { cin>>e;e--; mask|=(1<<(e)); } maskNumber.push_back(mask); } int ans=0,temp=0; for (int i = 0; i <n; i++) { for (int j = i+1; j < n; j++) { temp=__builtin_popcount( (maskNumber[i]) & (maskNumber[j])); ans=max(ans,temp); } } cout<<ans<<"\n"; }

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