April Fools Day Contest 2014--A

最新推荐文章于 2019-04-02 23:45:33 发布

转载最新推荐文章于 2019-04-02 23:45:33 发布 · 120 阅读

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原文链接：http://www.cnblogs.com/6bing/p/3931251.html

本文介绍了一个名为TheGreatGame的比赛，通过分析两个团队的动作字符串来判断胜负。输入为两行等长字符串，代表双方的动作（石头、剪刀、布），通过ASCII差值表示不同的动作。文章提供了一段C语言代码实现这一逻辑。

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Two teams meet in The Game World Championship. Some scientists consider this game to be the most intellectually challenging game in the world. You are given two strings describing the teams' actions in the final battle. Figure out who became the champion.

Input

The input contains two strings of equal length (between 2 and 20 characters, inclusive). Each line describes the actions of one team.

Output

Output "TEAM 1 WINS" if the first team won, "TEAM 2 WINS" if the second team won, and "TIE" if there was a tie.

Sample test(s)

Input

[]()[]8<
8<[]()8<

Output

TEAM 2 WINS

Input

8<8<()
[]8<[]

Output

TIE

意思：石头剪刀布（）代表石头（ASCII 差 1）， 8<代表剪刀（ASCII 差 4）， [ ] 代表布（ASCII 差 2）———比如： ‘ [ ’ 的ASCII 与 ‘ ] ’ ASCII 相差 2

#include<stdio.h>
#include<string.h>
char s1[100], s2[100];

int main()
{
	int i, j, count1=0, count2=0, len1 = 0, len2 = 0;
	gets(s1);
	gets(s2);
	len1 = strlen(s1);
	len2 = strlen(s2);
for(i=0; i<len1 -1; i+=2)
{
	if(s1[i+1] - s1[i] == 2 && s2[i+1] - s2[i] == 4  || s1[i+1] - s1[i] == 1 && s2[i+1] - s2[i] == 2 || s1[i+1] - s1[i] == 4 && s2[i+1] - s2[i] == 1)
		count2++;
	if(s1[i+1] - s1[i] == 2 && s2[i+1] - s2[i] == 1  || s1[i+1] - s1[i] == 1 && s2[i+1] - s2[i] == 4 || s1[i+1] - s1[i] == 4 && s2[i+1] - s2[i] == 2)
		count1++;
}
if(count1 == count2)
     printf("TIE\n");
if(count1 > count2)
     printf("TEAM 1 WINS\n");
if(count1 < count2)
     printf("TEAM 2 WINS\n");

	 return 0;
}

转载于:https://www.cnblogs.com/6bing/p/3931251.html