Codeforces 1176B. Merge it! Solution

Idea: Perform modulo 3 preprocessing on all elements, then greedy the pairing of remainder 1 and remainder 2, and the remaining 3 pairs are paired.

 Solution in C++:

///**********ALLAH IS ALMIGHTY************///

///AH Tonmoy

///Department of CSE,23rd batch

///Islamic University,Bangladesh  

1. #include<iostream>
2. using namespace std;
3. int main()
4. {
5. int t,n;
6. cin>>t;
7. while(t--)
8. {
9. cin>>n;
10. int one=0,two=0,three=0;
11. int a[n+3];
12. int i;
13. for(i=0; i<n; i++)
14. {
15. cin>>a[i];
16. if(a[i]%3==0)
17. three++;
18. else if(a[i]%3==1)
19. one++;
20. else
21. two++;
22. }
23. int x,y,x1,y1,mn;
24. mn=min(one,two);
25. x=one-mn;
26. y=two-mn;
27. x1=x/3;
28. y1=y/3;
29. cout<<three+mn+x1+y1<<endl;
30. }
31. }