[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：

10C JJ BC BB BB CC CC BJ BB CJ J

输出样例：

5 3 22 3 5B 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;}