【乱搞】【AOJ-34】Euchre Results

Description

Anna, Betty, Cindy and Zelda like playing the card game Euchre. Euchre is a game for two teams of two, and each time they meet the girls split off into different teams. They also keep overall records of the number of games each player has won and lost. Anna has misplaced her won-loss results, but she does have the results of the other three players. Given this, she figures she can determine her won-loss record.

 

Input

Input will consist of multiple problem instances. Each instance will consist of a single line containing six integers. The first two are the number of wins and losses (respectively) for Betty, the next two are the number of wins and losses for Cindy and the last two are the number of wins and losses for Zelda.
A final line of six zeroes will terminate input and should not be processed.

 

Output

For each problem instance, output a single line indicating Anna's won-loss record, in the format shown in the example below.

 

Sample Input

10 3 6 7 8 5
1874 2945 2030 2789 1025 3794
0 0 0 0 0 0

 

Sample Output

Anna's won-loss record is 2-11.
Anna's won-loss record is 4709-110.

大致翻译：四个女孩玩一种卡牌游戏，她们玩了很多场。两人一组，并且可以随时换组，比如这场AB一组 CD一组 ，AB赢了AB记一次win，BC输了BC记一次lose 下一局AC一组，BD一组 以此类推

现在Anna忘了她输赢多少局了，给出其他三个女孩的输赢次数 算出来Anna输赢多少局

 

思路：
解二元一次方程组即可

易知总的输赢场数相等，因为有赢必有输 设Anna win x lose y

则x+a+c+e=y+b+d+f

还有她们所有人玩的局数是相同的

则x+y=a+b=c+d=e+f

解出来即可

 

参考代码:

#include <stdio.h> 
int main() 
{ 
    int a,b,c,d,e,f; 
    while(scanf("%d%d%d%d%d%d",&a,&b,&c,&d,&e,&f)&&(a||b||c||d||e||f)) 
        printf("Anna's won-loss record is %d-%d.\n",(2*b+d+f-c-e)/2,(2*a+c+e-d-f)/2); 
    return 0; 
}