TheBoredomDivTwo
SRM 488 · 2010-03-12 · by Vasyl[alphacom]
SRM 488 · 2010-03-12 · by Vasyl[alphacom] · Simple Search, Iteration
Problem Statement
Problem Statement
John and Brus are bored.
They have n+m common friends.
The first n of them are bored and other m are not.
John chooses the j-th (1-based) friend for a talk.
If the friend is not bored, he becomes bored after the talk.
Brus does the same with the b-th (1-based) friend.
Note that John and Brus can't choose the same friend.
You have to find the number of bored friends after the talks.
You have to find the number of bored friends after the talks.
Constraints
- n will be between 1 and 47, inclusive.
- m will be between 1 and 47, inclusive.
- j will be between 1 and n+m, inclusive.
- b will be between 1 and n+m, inclusive.
- j and b will be different.
Examples
0)
1 1 1 2 Returns: 2
The first friend is already bored and the second friend becomes bored after the talk with Brus.
1)
2 1 1 2 Returns: 2
Here John and Brus choose two friends that are already bored.
2)
1 2 3 2 Returns: 3
All the friends become bored.
3)
4 7 7 4 Returns: 5
4)
9 9 18 13 Returns: 11
Submissions are judged against all 67 archived test cases, of which 5 are shown here. Case numbers match the judge’s.
Coding Area
Language: C++17 · define a public class TheBoredomDivTwo with a public method int find(int n, int m, int j, int b) · 67 test cases · 2 s / 256 MB per case