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Application of Modular Arithmetic

 

Q: If you have x,y,n. and you have to find a number k between 0<=n so that,
k%x=y


Approach1(Bruteforce):
1.Check if n%x==y print n
2.if n%x!=y then loop until n%x==y then print n. 

Code:
  1. #include <bits/stdc++.h>
  2. using namespace std;
  3. int main(){
  4.  
  5. int t;cin>>t;
  6. while(t--)
  7. {
  8. int x,y,n;cin>>x>>y>>n;
  10. if(n%x==y)cout<<n<<endl;
  11. else
  12. {
  13. while(n%x!=y)
  14. {
  15. n--;
  16. }
  17. cout<<n<<endl;
  18. }
  19. }
  20. }
Approach2(Greedy):
Time Complexity: O(n)

let assume, k=p'*x+y
now, p'=(k-y)/x ; 0<=k<=n;
so p'=floor((k-y)/x)
ans=k=p'*x+y 
Theory Review: 

Code:
  1. #include <bits/stdc++.h>
  2. using namespace std;
  3. int main(){
  4.  
  5. int t;cin>>t;
  6. while(t--)
  7. {
  8. int x,y,n;cin>>x>>y>>n;
  10. int p=floor(n-y)/x;
  11. cout<<p*x+y<<endl;
  12. }
  13. }


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