1018 錘子剪刀布 （20 分）
大家應該都會玩“錘子剪刀布”的遊戲：兩人同時給出手勢，勝負規則如圖所示：
FigCJB.jpg
現給出兩人的交鋒記錄，請統計雙方的勝、平、負次數，並且給出雙方分別出什麼手勢的勝算最大。
輸入格式：
輸入第 1 行給出正整數 N（≤105），即雙方交鋒的次數。隨後 N 行，每行給出一次交鋒的信息，即甲、乙雙方同時給出的的手勢。C 代表“錘子”、J 代表“剪刀”、B 代表“布”，第 1 個字母代表甲方，第 2 個代表乙方，中間有 1 個空格。
輸出格式：
輸出第 1、2 行分別給出甲、乙的勝、平、負次數，數字間以 1 個空格分隔。第 3 行給出兩個字母，分別代表甲、乙獲勝次數最多的手勢，中間有 1 個空格。如果解不唯一，則輸出按字母序最小的解。
輸入樣例：
10
C J
J B
C B
B B
B C
C C
C B
J B
B C
J J
輸出樣例：
5 3 2
2 3 5
B B
#include <cstdio>
#include <cstring>
#include <cmath>
int main()
{
int N = 0;
scanf("%d", &N);
int draw = 0;//平局次數
int win1 = 0, defeat1 = 0;//甲通過出🔨獲勝和失敗的次數
int win2 = 0, defeat2 = 0;//甲通過出✂獲勝和失敗的次數
int win3 = 0, defeat3 = 0;//甲通過出✋獲勝和失敗的次數
getchar();
for (int i = 0; i < N; i++)
{
char AA, BB;//表示甲乙出的手勢
//scanf_s("%c %c", &AA, &BB);
AA = getchar();
getchar();
BB = getchar();
getchar();
if (AA == 'C')//甲出🔨
{
if (BB == 'C') draw++;//乙也出🔨則平局
if (BB == 'J') win1++;//乙出✂則甲勝利
if (BB == 'B') defeat1++;//乙出✋則甲失敗
}
if (AA == 'J')//甲出✂
{
if (BB == 'C') defeat2++;//乙也出🔨則失敗
if (BB == 'J') draw++;//乙出✂則甲平局
if (BB == 'B') win2++;//乙出✋則甲勝利
}
if (AA == 'B')//甲出✋
{
if (BB == 'C') win3++;//乙也出🔨則勝利
if (BB == 'J') defeat3++;//乙出✂則甲失敗
if (BB == 'B') draw++;//乙出✋則甲平局
}
}
int win = win1 + win2 + win3;//甲獲勝總次數（或者說乙失敗總次數）
int defeat = defeat1 + defeat2 + defeat3;//甲失敗總次數（或者說乙勝利總次數）
printf("%d %d %d\n", win, draw, defeat);
printf("%d %d %d\n", defeat, draw, win);
if (win3 >= win1 && win3 > win2 || win3 > win1 && win3 >= win2) printf("B");//甲通過出✋獲勝次數最多
else if (win1 >= win2 && win1 > win3 || win1 > win2 && win1 >= win3) printf("C");//甲通過出🔨獲勝次數最多
else if (win2 >= win1 && win2 > win3 || win2 > win1 && win2 >= win3) printf("J");//甲通過出✂獲勝次數最多
else printf("B");//否則出✋；如果解不唯一，則輸出按字母序最小的解。
if (defeat1 >= defeat2 && defeat1 > defeat3 || defeat1 > defeat2 && defeat1 >= defeat3) printf(" B\n");
else if (defeat2 >= defeat1 && defeat2 > defeat3 || defeat2 > defeat1 && defeat2 >= defeat3) printf(" C\n");
else if (defeat3 >= defeat1 && defeat3 > defeat2 || defeat3 > defeat1 && defeat3 >= defeat2) printf(" J\n");
else printf(" B\n");
return 0;
}