,即双方交锋的次数。随后N行,每行给出一次交锋的信息,即甲、乙双方同时给出的的手势。C代表“锤子”、J代表“剪刀”、B代表“布”,第1个字母代表甲方,第2个代表乙方,中间有1个空格。

#include<iostream>
using namespace std;
int main()
{
	char MaxNum(int a, int b, int c);
	/*
	 * 输入格式:

输入第1行给出正整数N(<=105),即双方交锋的次数。随后N行,每行给出一次交锋的信息,即甲、乙双方同时给出的的手势。C代表“锤子”、J代表“剪刀”、B代表“布”,第1个字母代表甲方,第2个代表乙方,中间有1个空格。

输出格式:

输出第1、2行分别给出甲、乙的胜、平、负次数,数字间以1个空格分隔。第3行给出两个字母,分别代表甲、乙获胜次数最多的手势,中间有1个空格。如果解不唯一,则输出按字母序最小的解。

输入样例:
10
C J
J B
C B
B B
B C
C C
C B
J B
B C
J J
输出样例:
5 3 2
2 3 5
B B
	 */

	long count = 0;
	cin >> count;
	long arrAwin[3] = { 0 };
	long countNumA[3] = { 0 };//B C J
	long countNumB[3] = { 0 };//B C J
	for (size_t i = 0; i < count; i++)
	{
		char A;
		char B;
		cin >> A;
		cin >> B;
		if ((A == 'C'&&B == 'J') || (A == 'J'&&B == 'B') || (A == 'B'&&B == 'C'))
		{
			arrAwin[0]++;
			if (A == 'B')
			{
				countNumA[0]++;
			}
			if (A == 'C')
			{
				countNumA[1]++;
			}
			if (A == 'J')
			{
				countNumA[2]++;
			}
		}
		//if ((A == 'C'&&B == 'C') || (A == 'J'&&B == 'J') || (A == 'B'&&B == 'B'))
		if (A == B)
		{
			arrAwin[1]++;
		}
		if ((A == 'J'&&B == 'C') || (A == 'B'&&B == 'J') || (A == 'C'&&B == 'B'))
		{
			arrAwin[2]++;
			if (B == 'B')
			{
				countNumB[0]++;
			}
			if (B == 'C')
			{
				countNumB[1]++;
			}
			if (B == 'J')
			{
				countNumB[2]++;
			}
		}
			
	}
	cout << arrAwin[0] << " " << arrAwin[1] << " " << arrAwin[2] << endl;
	cout << arrAwin[2] << " " << arrAwin[1] << " " << arrAwin[0] << endl;
	cout << MaxNum(countNumA[0], countNumA[1], countNumA[2]) << " " << MaxNum(countNumB[0], countNumB[1], countNumB[2]) << endl;
	return 0;

}
char MaxNum(int a, int b, int c)
{
	if (a >= b&&a >= c)
	{
		return 'B';
	}
	if (b >= a&&b >= c)
	{
		return 'C';
	}
	if (c >= a&&c >= b)
	{
		return 'J';
	}
}

  

原文地址:https://www.cnblogs.com/linxuemufeng/p/8886767.html