PAT- 锤子剪刀布 (20)

版权声明:一只努力变强的编程汪,欢迎交流! https://blog.csdn.net/qq_42020563/article/details/81568456

锤子剪刀布 (20)

  •  
  •  
  •  
  •  

时间限制 1000 ms 内存限制 32768 KB 代码长度限制 100 KB 判断程序 Standard (来自 小小)

题目描述

大家应该都会玩“锤子剪刀布”的游戏:

现给出两人的交锋记录,请统计双方的胜、平、负次数,并且给出双方分别出什么手势的胜算最大。

输入描述:

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。


 

输出描述:

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。

输入例子:

10

C J

J B

C B

B B

B C

C C

C B

J B

B C

J J

输出例子:

5 3 2

2 3 5

B B
#include<iostream>
using namespace std;
int main()
{
	int jiaw = 0, jiap = 0, jiaf = 0, yiw = 0, yip = 0, yif = 0;
	int jiac = 0, jiaj = 0, jiab = 0, yic = 0, yij = 0, yib = 0;
	char jia, yi;
	int n;
	cin >> n;
	for (int i = 0; i<n; i++)
	{
		cin >> jia >> yi;
		if (jia == 'C'&&yi == 'J')
		{
			jiac++;
			jiaw++;
			yif++;
		}
		else if (jia == 'J'&&yi == 'B')
		{
			jiaj++;
			jiaw++;
			yif++;
		}
		else if (jia == 'B'&&yi == 'C')
		{
			jiab++;
			jiaw++;
			yif++;
		}
		else if (jia == yi)
		{
			jiap++;
			yip++;
		}
		else if (jia == 'C'&&yi == 'B')
		{
			yiw++;
			yib++;
			jiaf++;
		}
		else if (jia == 'B'&&yi =='J')
		{
			yiw++;
			yij++;
			jiaf++;
		}
		else if (jia == 'J'&&yi == 'C')
		{
			yiw++;
			yic++;
			jiaf++;
		}
	}
	cout << jiaw << " " << jiap << " " << jiaf << endl;
	cout << yiw << " " << yip << " " << yif << endl;
	if (jiac == 0 && jiab == 0 && jiaj == 0)
	{
		cout << "B ";
	}
	else
	{
		if (jiac > jiab)
		{
			if (jiac > jiaj)
				cout << "C ";
			else if (jiac<jiaj)
				cout << "J ";
			else if (jiac == jiaj)
				cout << "C ";
		}
		else if (jiab>jiac)
		{
			if (jiab > jiaj)
				cout << "B ";
			else if (jiab < jiaj)
				cout << "J ";
			else if (jiab == jiaj)
				cout << "B ";
		}
		else if (jiab == jiac)
		{
			if (jiab<jiaj)
				cout << "J ";
			else
				cout << "B ";
		}
	}
	if (yib == 0 && yij == 0 && yic == 0)
		cout << "B";
	else
	{
		if (yic>yib)
		{
			if (yic > yij)
				cout << "C";
			else if (yic<yij)
				cout << "J";
			else if (yic == yij)
				cout << "C";
		}
		else if (yib>yic)
		{
			if (yib > yij)
				cout << "B";
			else if (yib < yij)
				cout << "J";
			else if (yib == yij)
				cout << "B";
		}
		else if (yib == yic)
		{
			if (yib < yij)
				cout << "J";
			else
				cout << "B";
		}
	}
	return 0;
}

猜你喜欢

转载自blog.csdn.net/qq_42020563/article/details/81568456
今日推荐