题目链接:

https://pintia.cn/problem-sets/994805260223102976/problems/994805304020025344

我的代码:

#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <map>
using namespace std;


typedef pair<char, int> PAIR;
ostream& operator<<(ostream& out, const PAIR& p) {
	return out << p.second << "\t" << p.second;
}

int main() {
	int n;
	cin >> n;

	int result[3] = { 0 };
	int C_A = 0, B_A = 0, J_A = 0;
	int C_B = 0, B_B = 0, J_B = 0;
	//result[0]，甲赢,
	//result[1]，平局
	//result[2]，乙赢
	while (n--) {
		char a, b;

		cin >> a >> b;//a--甲方,b--乙方
		if (a == 'C'&&b == 'J') {
			result[0]++;
			C_A++;
		}
		else if (a == 'C'&&b == 'B') {
			result[2]++;
			B_B++;
		}
		else if (a == 'B'&&b == 'C') {
			result[0]++;
			B_A++;
		}
		else if (a == 'B'&&b == 'J') {
			result[2]++;
			J_B++;
		}
		else if (a == 'J'&&b == 'C') {
			result[2]++;
			C_B++;
		}
		else if (a == 'J'&&b == 'B') {
			result[0]++;
			J_A++;
		}
		else {
			result[1]++;
		}
		////输出次数
		//char max_a;
		//int max_A = max(max(B_A,C_A), J_A);

	}
	cout << result[0] << " " << result[1] << " " << result[2] << endl;
	cout << result[2] << " " << result[1] << " " << result[0] << endl;
	map<char, int,less<int>> A_{ {'B',B_A},{'C',C_A},{'J',J_A} };
	map<char, int,less<int>> B_{ {'B',B_B},{'C',C_B},{'J',J_B} };
	map<char, int, less<int>>::iterator it_a = A_.begin();
	map<char, int, less<int>>::iterator it_b = A_.begin();
	cout << it_a->first << " " << it_b->first ;
}

AC代码(参考 算法笔记):

#include <iostream>
#include <cstdio>

using namespace std;

int change(char c){
    if(c=='B')  return 0;
    if(c=='C')  return 1;
    if(c=='J')  return 2;
}

const int maxn=1000001;


int main(){
    int N;
    char mp[3]={'B','C','J'};
    int time_[3]={0};//0：代表甲赢,1：代表平局,2：代表乙赢
    int hard_A[3]={0},hard_B[3]={0};//分别代表甲,乙获胜次数最多的手势
    scanf("%d",&N);
    while(N--){
        int k1,k2;
        char c1,c2;
        getchar();//由于scanf使用%c时会将换行符\n读入,因此需要在合适的地方用getchar吸收空格
                    //否则,会导致程序读入数据后闪退,
        scanf("%c %c",&c1,&c2);
        k1=change(c1);
        k2=change(c2);
        if((k1+1)%3==k2){//甲赢
            time_[0]++;
            hard_A[k1]++;
        }
        else if((k2+1)%3==k1){//乙赢
            time_[2]++;
            hard_B[k2]++;
        }
        else{
            time_[1]++;
        }
    }
    printf("%d %d %d\n",time_[0],time_[1],time_[2]);
    printf("%d %d %d\n",time_[2],time_[1],time_[0]);

    int max_a=0,max_b=0;
    for(int i=1;i<3;i++){
        if(hard_A[max_a]<hard_A[i])
            max_a=i;
        if(hard_B[max_b]<hard_B[i])
            max_b=i;
    }
    printf("%c %c",mp[max_a],mp[max_b]);
    return 0;
}