PTA 1018 锤子剪刀布 (c语言)

1018 锤子剪刀布 (20 分)

这个题，我做的时候一直以为是有捷径可循的，但是后来怎么想都想不出来。
只能暴力了，所谓暴力出奇迹，哈哈哈 代码如下：

#include<stdio.h>
int main()
{
	int x,i,win=0,pin=0,lose=0,bu=0,ch=0,ji=0,bu1=0,ch1=0,ji1=0;
	if(scanf("%d",&x));
	char a[x],b[x];
	for(i=0;i<x;i++)
	{
		getchar();
		if(scanf("%c %c",&a[i],&b[i]));
		if((a[i]=='B'&&b[i]=='C')||(a[i]=='C'&&b[i]=='J')||(a[i]=='J'&&b[i]=='B'))
		{
			win++;
			if(a[i]=='B')
				bu++;
			else if(a[i]=='C')
				ch++;
			else
				ji++;
		}
		if((a[i]=='B'&&b[i]=='J')||(a[i]=='J'&&b[i]=='C')||(a[i]=='C'&&b[i]=='B'))
		{
			lose++;
			if(b[i]=='J')
				ji1++;
			else if(b[i]=='C')
				ch1++;
			else if(b[i]=='B')
				bu1++;
		}
		if(a[i]==b[i])
			pin++;
	}
	printf("%d %d %d\n",win,pin,lose);
	printf("%d %d %d\n",lose,pin,win);
	if((bu==ch&&ch==ji)||(bu==ch&&bu>ji)||(bu==ji&&bu>ch)||(bu>ch&&bu>ji))
		printf("B ");
	else if((ch==ji&&ch>bu)||(ch>ji&&ch>bu))
		printf("C ");
	else if(ji>bu&&ji>ch)
		printf("J ");
	if((bu1==ch1&&ch1==ji1)||(bu1==ch1&&bu1>ji1)||(bu1==ji1&&bu1>ch1)||(bu1>ch1&&bu1>ji1))
		printf("B");
	else if((ch1==ji1&&ch1>bu1)||(ch1>ji1&&ch1>bu1))
		printf("C");
	else if(ji1>bu1&&ji1>ch1)
		printf("J");
	return 0;	
}

我一直在纠结的是这个 输出它是靠什么赢得最多 这个点，所以我每次都会去判断它是什么手势，它的结果如何。
额。。。下面分享一个大神的代码。。。大神果然不一样https://www.liuchuo.net/archives/539
代码如下：

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;
}

来源：https://blog.csdn.net/qq_44622401/article/details/99047718