Building-would is ancient Chinese wine culture as part of a fun, two people on the wine table building-would of method is: in the mouth a number, each at the same time to her out of a number, if who holds a number equal to the sum of two people shout out of digital, who lose, loser for a cup of wine, two people to win or to lose two people continues to the next round, until the only winner,

Given below a, b two people's capacity for liquor (at most) can drink how much cup and building-would records, would you please determine two men who first down,

Input format:
Enter the first line has given a, b two people drink of the nonnegative integers (no more than 100), separated by Spaces, the next row is given a positive integer N (100) or less, then N lines, each line give a round of building-would records, format for:

A shout a cross b shout b row
Among them is shout shout out Numbers, row is drawn Numbers, are all positive integers less than 100 (two hands row),

The output format:
Output in the first row, first the man fall: on behalf of A and B on behalf of B, the second line of the output the man standing drink how much cup, ensure that there is A person fall, pay attention to the process end, someone fell behind the data don't have to deal with,

Input:
1 1
6
8 9 10 12
5 to 10 10
3 August 5 12
12 18 1 13
4 16 12 15
15 1 1 16

Output:
A
1

Please ask how this programming problem!
With the method of the Python!!!!!

CodePudding user response:

 result=[[0, False, 'A'], [0, False, 'B']] 
With the open (r 'c: \ ZZZ. TXT', encoding="utf-8", mode='r +') as f: 
L=f.r ead (.) splitlines () 
A, B=map (int, l [0]. The split ()) 
The count=int (l [1]) 
For I in range (count) : 
Aj, Ah, Bj, Bh=map (int, l [2 + I]. The split ()) 
If Aj + Bj Bj====Ah and Aj + Bh: 
The continue 
Elif Aj + Bj!=Ah and Aj + Bj!=Bh: 
The continue 
Elif Aj + Bj==Ah and Aj + Bj!=Bh: 
The result [0] [0] +=1 
The else: 
The result [1] [0] +=1 
If the result [0] [0] & gt; A: 
The result [0] [1]=True 
Break 
If the result [1] [0] & gt; B: 
The result [1] [1]=True 
Break 

Print (STR (list (filter (lambda x: x [1], the result)) [0] [2])) 
Print (STR (list (filter (lambda x: not x [1], the result)) [0] [0])) 

This mean?

CodePudding user response:

 
A_limit b_limit=[int (x) for x input in (). The split ()] 
N=int (input ()) 
A_count=0 
B_count=0 
For I in range (n) : 
TMP=[int (x) for x input in (). The split ()] 
If TMP TMP [0] + [2]==TMP [1] and TMP [0] + TMP [2].=TMP [3] : 
A_count +=1 
Elif TMP + TMP [0] [2].=TMP [1] and TMP [0] + TMP [2]==TMP [3] : 
B_count +=1 
If a_count & gt; A_limit: 
Print (' A ') 
Print (b_count) 
Break 
Elif b_count & gt; B_count: 
Print (' B ') 
Print (a_count) 
Break 

Probably like this

CodePudding user response:

Okok, done, thank you

CodePudding user response:

Quote: refer to 1st floor chuifengde response:

, but I haven't learned the files here, and I have some don't understand your answer
But you can make to
Thank you ~