我的北大ACM POJ 1013解答

我的北大ACM POJ 1013解答
又一个水题，居然还贡献了3次WA。。。

Source Code

Problem: 1013		User: absolute
Memory: 204K		Time: 0MS
Language: C++		Result: Accepted

• Source Code

#include <stdio.h>
#include <string>
void POJ1013();
int main()
{
	POJ1013();
	return 0;
}
void POJ1013()
{
	int i,k,casen;
	scanf("%d",&casen);
	while(casen>0)
	{
		casen--;
		bool real[12]={0};
		int heavy[12]={0};
		int light[12]={0};
		for(k=0;k<3;k++)
		{
			char input[3][8]={0};
			int j;
			scanf("%s %s %s",&input[0],&input[1],&input[2]);
			//平衡则放入真数组，表示这次称的都是真币
			if(strcmp(input[2],"even")==0)
			{
				j=0;
				while(input[0][j]!='\0')
				{
					real[input[0][j]-'A']=true;
					real[input[1][j]-'A']=true;
					j++;
				}
			}
			//右边起来，左边怀疑为重，右边怀疑为轻
			else if(strcmp(input[2],"up")==0)
			{
				j=0;
				while(input[0][j]!='\0')
				{
					heavy[input[0][j]-'A']++;
					light[input[1][j]-'A']++;
					j++;
				}
			}
			//右边下降，左边怀疑为轻，右边为重
			else if(strcmp(input[2],"down")==0)
			{
				j=0;
				while(input[0][j]!='\0')
				{
					light[input[0][j]-'A']++;
					heavy[input[1][j]-'A']++;
					j++;
				}

			}
		}
		char maxnum='A';
		int heavyorlight=0;
		//假币必然是不在真数组中，而且它必然只能在重数组和轻数组之一出现，而不能同时出现
		//但是有可能被怀疑的只称过一次，所以必须选出被怀疑最多次的那个
		for(i='A';i<'M';i++)
		{
			if(real[i-'A'])
				continue;
			int temp=0;
			if(light[i-'A']>0&&heavy[i-'A']==0)
			{
				if(heavyorlight==1)
					temp=light[maxnum-'A'];
				else if(heavyorlight==2)
					temp=heavy[maxnum-'A'];
				if(light[i-'A']>temp)
				{
					heavyorlight=1;
					maxnum=i;
				}
				continue;
			}
			if(light[i-'A']==0&&heavy[i-'A']>0)
			{
				if(heavyorlight==1)
					temp=light[maxnum-'A'];
				else if(heavyorlight==2)
					temp=heavy[maxnum-'A'];
				if(heavy[i-'A']>temp)
				{
					heavyorlight=2;
					maxnum=i;
				}
				continue;
			}
		}
		if(heavyorlight==1)
			printf("%c is the counterfeit coin and it is light.\n",maxnum);
		else
			printf("%c is the counterfeit coin and it is heavy.\n",maxnum);
	}
}