[PAT-乙级]1018.锤子剪刀布

1018. 锤子剪刀布 (20)

时间限制

100 ms

内存限制

65536 kB

代码长度限制

8000 B

判题程序

Standard

作者

CHEN, Yue

大家应该都会玩“锤子剪刀布”的游戏：两人同时给出手势，胜负规则如图所示：



现给出两人的交锋记录，请统计双方的胜、平、负次数，并且给出双方分别出什么手势的胜算最大。

输入格式：

输入第1行给出正整数N（<=105），即双方交锋的次数。随后N行，每行给出一次交锋的信息，即甲、乙双方同时给出的的手势。C代表“锤子”、J代表“剪刀”、B代表“布”，第1个字母代表甲方，第2个代表乙方，中间有1个空格。

输出格式：

输出第1、2行分别给出甲、乙的胜、平、负次数，数字间以1个空格分隔。第3行给出两个字母，分别代表甲、乙获胜次数最多的手势，中间有1个空格。如果解不唯一，则输出按字母序最小的解。
输入样例：

10
C J
J B
C B
B B
B C
C C
C B
J B
B C
J J

输出样例：

5 3 2
2 3 5
B B

甲赢得次数其实与乙输的次数相等，甲乙平手的次数相等

#include<stdio.h>
#include<string.h>

int main()
{
//freopen("D://input.txt", "r", stdin);
char a[2], b[2];
int n;
while(scanf("%d", &n) != EOF)
{
int count_a[6];//count_a1中角码0表示甲胜的次数，1->甲平的次数，2->甲输的次数，3->甲赢出锤子，4->甲赢出布，5->甲赢出剪刀
int count_b[6];
memset(count_a, 0, sizeof(count_a));
memset(count_b, 0, sizeof(count_b));
for(int i = 0; i < n; i ++)
{

scanf("%s %s", a, b);
if((a[0] == 'C' && b[0] == 'C') || (a[0] == 'J' && b[0] == 'J') || (a[0] == 'B' && b[0] == 'B'))
{
count_a[1] ++;
count_b[1] ++;
}
else if((a[0] == 'B' && b[0] == 'C') || (a[0] == 'C' && b[0] == 'J') || (a[0] == 'J' && b[0] == 'B'))
{
count_a[0] ++;
count_b[2] ++;
if(a[0] == 'C')
count_a[3] ++;
else if(a[0] == 'B')
count_a[4] ++;
else
count_a[5] ++;
}
else
{
count_a[2] ++;
count_b[0] ++;

if(b[0] == 'C')
count_b[3] ++;
else if(b[0] == 'B')
count_b[4] ++;
else
count_b[5] ++;
}
}
printf("%d %d %d\n", count_a[0], count_a[1], count_a[2]);
printf("%d %d %d\n", count_b[0], count_b[1], count_b[2]);

if(count_a[4] >= count_a[3] && count_a[4] >= count_a[5])
printf("B ");
else if(count_a[3] >= count_a[4] && count_a[3] >= count_a[5])
printf("C ");
else if(count_a[5] >= count_a[3] && count_a[5] >= count_a[4])
printf("J ");

if(count_b[4] >= count_b[3] && count_b[4] >= count_b[5])
printf("B\n");
else if(count_b[3] >= count_b[4] && count_b[3] >= count_b[5])
printf("C\n");
else if(count_b[5] >= count_b[3] && count_b[5] >= count_b[4])
printf("J\n");
}
return 0;
}