【PAT乙级】 1018 锤子剪刀布 (20 分)

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该博客包含两种解法来统计和分析锤子剪刀布游戏的结果。解法一和解法二分别通过不同方式计算平局、甲方胜利和乙方胜利的次数，并输出胜率最高的出拳选项。

1018锤子剪刀布 (20分)

//解法一
#include <iostream>

using namespace std;

int n, cnt, sumA, sumB;
int a[3], b[3];

int get(string s)
{
    if (s == "J") return 0;
    if (s == "C") return 1;
    if (s == "B") return 2;
}

string uget(int t)
{
    if (t == 0) return "J";
    if (t == 1) return "C";
    if (t == 2) return "B";
}

int getmax(int a[])
{
    int x = 2;
    for (int i = 1; i >= 0; i --)
        if (a[i] > a[x]) x = i;
    return x;
}

int main()
{
    cin >> n;
    for (int i = 0; i < n; i ++)
    {
        string x, y;
        cin >> x >> y;
        if (get(x) == get(y)) cnt ++;
        else if ((get(x) + 1) % 3 == get(y)) b[get(y)]++;
        else a[get(x)]++;
    }
    
    for (int i = 0; i <= 2; i ++)
    {
        sumA += a[i];
        sumB += b[i];
    }

    cout << sumA << ' ' << cnt << ' ' << sumB << endl;
    cout << sumB << ' ' << cnt << ' ' << sumA << endl;

    cout << uget(getmax(a)) << ' ' << uget(getmax(b)) << endl;
    return 0;
}

//解法二
#include<iostream>

using namespace std;

int n;
int jiawin,yiwin, ping;
int jiages[3];
int yiges[3];

int get(char c)
{
    if (c == 'c') return 0;
    if (c == 'j') return 1;
    return 2;
}
int main()
{
    cin >> n;
    for (int i = 0; i < n; i ++)
    {
        char a, b;
        cin >> a >> b;
        int x = get(a), y = get(b);
        if (x == y) 
        {
            ping ++;            
        }
        else if ((x + 1) % 3 == y)
        {
            jiawin ++;
            jiages[x] ++;
        }
        else
        {
            yiwin ++;
            yiges[y] ++;
        }
    }
    

    cout << jiawin <<" "<< n - jiawin - yiwin <<" " << yiwin << endl;
    cout << yiwin <<" "<< n - jiawin - yiwin <<" "<< jiawin <<endl;

    char maxges[4] = {"BCJ"};
    int maxjia = jiages[0] >= jiages[1] ? 0 : 1;
    maxjia = jiages[maxjia] >= jiages[2] ? maxjia : 2;

    int maxyi = yiges[0] >= yiges[1] ? 0 : 1;
    maxyi = yiges[maxjia] >= yiges[2] ? maxyi : 2;

    cout << maxges[maxjia] <<" "<<maxges[maxyi];

    return 0;
}