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要解决这个问题，我们需要统计“锤子剪刀布”游戏中双方的胜、平、负次数，并找出双方分别出什么手势的胜算最大。

方法思路

解决代码

n = int(input())jia_arr = [0, 0, 0]  # B, C, Jyi_arr = [0, 0, 0]win_jia = 0ping_jia = 0fu_jia = 0for _ in range(n):    jia, yi = input().split()    if jia == yi:        ping_jia += 1    else:        if (jia == 'C' and yi == 'J') or (jia == 'J' and yi == 'C'):            win_jia += 1            jia_arr[1] += 1        elif (jia == 'C' and yi == 'B') or (jia == 'B' and yi == 'C'):            fu_jia += 1            jia_arr[0] += 1        elif (jia == 'J' and yi == 'B') or (jia == 'B' and yi == 'J'):            fu_jia += 1            jia_arr[2] += 1        elif (jia == 'J' and yi == 'C') or (jia == 'C' and yi == 'J'):            win_jia += 1            jia_arr[2] += 1        elif (jia == 'B' and yi == 'J') or (jia == 'J' and yi == 'B'):            win_jia += 1            jia_arr[0] += 1        elif (jia == 'B' and yi == 'C') or (jia == 'C' and yi == 'B'):            fu_jia += 1            jia_arr[1] += 1def get_max(arr):    max_val = -1    max_index = 0    for i in range(len(arr)):        if arr[i] > max_val:            max_val = arr[i]            max_index = i    return arr[max_index]flag1 = get_max(jia_arr)flag2 = get_max(yi_arr)print(f"{win_jia} {ping_jia} {fu_jia}")print(f"{fu_jia} {ping_jia} {win_jia}")print(f"{flag1} {flag2}")

代码解释

这个方法确保了代码的简洁和逻辑的清晰，能够正确统计双方的胜负次数并确定胜利手势。

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