[刷题] 1018 锤子剪刀布 (20分)

思路

• jiawin、yiwin分别表示甲乙赢的次数
• s和t分别表示每一次甲乙给出的手势
• jia、yi记录获胜手势的次数
• maxjia和maxyi分别表示甲乙获胜次数最多的手势所对应的下标（012分别表示BCJ）
• 枚举每一次甲乙手势的胜负结果，累加到jiawin和yiwin中，输出结果

#include <iostream>
using namespace std;
int main() {
    int n;
    cin >> n;
    int jiawin = 0, yiwin = 0;
    int jia[3] = {0}, yi[3] = {0};
    for (int i = 0; i < n; i++) {
        char s, t;
        cin >> s >> t;
        if (s == 'B' && t == 'C') {
            jiawin++;
            jia[0]++;
        } else if (s == 'B' && t == 'J') {
            yiwin++;
            yi[2]++;
        } else if (s == 'C' && t == 'B') {
            yiwin++;
            yi[0]++;
        } else if (s == 'C' && t == 'J') {
            jiawin++;
            jia[1]++;
        } else if (s == 'J' && t == 'B') {
            jiawin++;
            jia[2]++;
        } else if (s == 'J' && t == 'C') {
            yiwin++;
            yi[1]++;
        }
    }
    cout << jiawin << " " << n - jiawin - yiwin << " " << yiwin << endl << yiwin << " " << n - jiawin - yiwin << " " << jiawin << endl;
    int maxjia = jia[0] >= jia[1] ? 0 : 1;
    maxjia = jia[maxjia] >= jia[2] ? maxjia : 2;
    int maxyi = yi[0] >= yi[1] ? 0 : 1;
    maxyi = yi[maxyi] >= yi[2] ? maxyi : 2;
    char str[4] = {"BCJ"};
    cout << str[maxjia] << " " << str[maxyi];
    return 0;
}

参考

https://www.liuchuo.net/archives/539