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1 A J up B J up C J up, 答案一目了然吧

Posted by xfxyjwf at 2006-05-18 09:09:53 on Problem 1013
In Reply To:路过的给看一下 Posted by:lzu311 at 2006-05-18 08:44:41
推理很容易出错,太容易漏点什么了
觉得枚举是个不错的办法:)

> #include<stdio.h>
> #include<string.h>
> void main()
> {
> 	int i,n,j,k;
> 	char s1[10],s2[10],s3[10];
> 	int s[20];
> 	scanf("%d",&n);
> 	for(i=1;i<=n;i++)
> 	{
> 		memset(s,0,sizeof(s));
> 		for(j=1;j<=3;j++)
> 		{
> 			scanf("%s%s%s",s1,s2,s3);
> 			if(strcmp(s3,"even")==0)//如果even,置为1,其为真币
> 			{
> 				for(k=0;s1[k]!=0;k++)
> 				{
> 					s[s1[k]-'A']=1;
> 					s[s2[k]-'A']=1;
> 				}
> 			}
> 			if(strcmp(s3,"up")==0)
> 			{
> 				for(k=0;s1[k]!=0;k++)
> 				{
> 					if(s[s2[k]-'A']!=1)
> 					{
> 						if(s[s2[k]-'A']==2)//存在矛盾,为真币
> 							s[s2[k]-'A']=1;
> 						else
> 							s[s2[k]-'A']=-1;//-1标识假币为light
> 					}
> 					if(s[s1[k]-'A']!=1)
> 					{
> 						if(s[s1[k]-'A']==-1)
> 							s[s1[k]-'A']=1;
> 						else
> 							s[s1[k]-'A']=2;//2标识假币为heavy
> 					}
> 
> 				}
> 			}
> 			if(strcmp(s3,"down")==0)
> 			{
> 				for(k=0;s2[k]!=0;k++)
> 				{
> 					if(s[s2[k]-'A']!=1)
> 					{
> 						if(s[s2[k]-'A']==-1)
> 							s[s2[k]-'A']=1;
> 						else
> 							s[s2[k]-'A']=2;
> 					}
> 					if(s[s1[k]-'A']!=1)
> 					{
> 						if(s[s1[k]-'A']==2)
> 							s[s1[k]-'A']=1;
> 						else
> 							s[s1[k]-'A']=-1;
> 					}
> 				}
> 			}
> 		}
> 		for(j=0;j<20;j++)
> 		{
> 			if(s[j]==-1)
> 			{
> 				printf("%c is the counterfeit coin and it is light.\n",j+'A');
> 				break;
> 			}
> 			if(s[j]==2)
> 			{
> 				printf("%c is the counterfeit coin and it is heavy.\n",j+'A');
> 				break;
> 			}
> 		}
> 	}
> }

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