1018 锤子剪刀布 (20 分)（附注释，逻辑分析）

写在前面

• 使用整型变量、字符定义所有值

• 冗余，繁琐，易出错
• 空格属于字符，读取处理比较麻烦

• map 封装，计算最优手势
• 计数甲、乙生产手势
• 题目不难，输入处理耗费部分时间

• 普通水平，10分钟内a掉

测试用例

input: 
10
C J
J B
C B
B B
B C
C C
C B
J B
B C
J J

output: 
5 3 2
2 3 5

ac代码

• 个人代码（推荐，处理字符串）

#include <map>
#include <iostream>
using namespace std;

int main()
{
    int n;
    scanf("%d\n", &n);

    int js=0, ys=0;
    string bcj;
    map<char,int> jm, ym;

    for(int i=0; i<n; i++)
    {
        getline(cin, bcj);
        if(((bcj[0]=='C')&&(bcj[2]=='J'))||((bcj[0]=='J')&&(bcj[2]=='B'))||((bcj[0]=='B')&&(bcj[2]=='C')))
        {
            js++;
            jm[bcj[0]] += 1;
        }
        if(((bcj[0]=='J')&&(bcj[2]=='C'))||((bcj[0]=='B')&&(bcj[2]=='J'))||((bcj[0]=='C')&&(bcj[2]=='B')))
        {
            ys++;
            ym[bcj[2]] += 1;
        }
    }

//    printf("%d %d %d\n", jm['B'], jm['C'], jm['J']);
//    printf("%d %d %d\n", ym['B'], ym['C'], ym['J']);

    printf("%d %d %d\n", js, n-ys-js, ys);
    printf("%d %d %d\n", ys, n-ys-js, js);
    printf("%c %c", jm['B']>=jm['C'] ? (jm['B']>=jm['J'] ? 'B' : 'J') : (jm['C']>=jm['J'] ? 'C' : 'J'),ym['B']>=ym['C'] ? (ym['B']>=ym['J'] ? 'B' : 'J') : (ym['C']>=ym['J'] ? 'C' : 'J'));

    return 0;
}

• ​​参考代码（处理字符）​​

#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;
}