Hrbeu stone scissors cloth 1

Stone scissors cloth 1

Timelimit: 1 second memorylimit: 32
Megabyte

Totalsubmit: 73 accepted: 30

Description

After winter vacation, X is bored at home. So he came up with a new stone scissors cloth (single-host ).

The method is as follows: S represents stone, j Represents scissors, and B Represents cloth. If you win one point, you will not be scored. If you lose one point, you will be deducted. Play n times, and each time the opponent plays a game, everything is known. However, you can only output a s, B j, and c B (A, B, C are both non-negative and A + B + C = N ).

Now, how do you arrange for a S, B j, and c B to maximize your final score.

Input

N (n <= 100) is given first, indicating the number of games. Next, we will give n strings consisting of S, J, and B to indicate what the opponent is after each game. Finally, A, B, and C are given.

When n = 0, the input ends.

Output

An integer is given for each group of test data, indicating the maximum score that X can win in this game.

Sample Input

2

Jj

2 0 0

0

Sample output

2

Source

Author: xiaoz

 

# Include <cstdio>
# Include <cmath>
# Include <algorithm>
Using namespace STD;
Int N, A1, A2, B1, B2, C1, C2, sum = 0, sum2 = 0;
Char ch;
Int main (){
While (~ Scanf ("% d", & N), n ){
A1 = b1 = C1 = 0;
Getchar ();
While (CH = getchar ())! = '\ N '){
If (CH = 'J') B1 ++;
If (CH = 's') A1 ++;
If (CH = 'B') C1 ++;
}
Scanf ("% d", & A2, & B2, & C2 );
Sum = min (A2, B1) + min (B2, C1) + min (C2, A1 );
If (a2-b1> 0 ){
A2 = a2-b1;
B1 = 0;
}
Else {
B1 = b1-a2;
A2 = 0;
}
If (b2-c1> 0 ){
B2 = b2-c1;
C1 = 0;
}
Else {
C1 = c1-b2;
B2 = 0;
}
If (c2-a1> 0 ){
C2 = c2-a1;
A1 = 0;
}
Else {
A1 = a1-c2;
C2 = 0;
}
Sum2 = min (a1, a2) + min (b1, b2) + min (C1, C2 );
Sum = sum-(n-sum2-sum );
Printf ("% d \ n", sum );
}
Return 0;
}