poj-Counterfeit Dollar

 

描述

Sally Jones has a dozen Voyageur silver dollars. However, only eleven of the coins are true silver dollars; one coin is counterfeit even though its color and size make it indistinguishable from the real silver dollars. The counterfeit coin has a different weight from the other coins but Sally does not know if it is heavier or lighter than the real coins. 

Happily, Sally has a friend who loans her a very accurate balance scale. The friend will permit Sally three weighings to find the counterfeit coin. For instance, if Sally weighs two coins against each other and the scales balance then she knows these two coins are true. Now if Sally weighs 
one of the true coins against a third coin and the scales do not balance then Sally knows the third coin is counterfeit and she can tell whether it is light or heavy depending on whether the balance on which it is placed goes up or down, respectively. 

By choosing her weighings carefully, Sally is able to ensure that she will find the counterfeit coin with exactly three weighings.

输入

The first line of input is an integer n (n > 0) specifying the number of cases to follow. Each case consists of three lines of input, one for each weighing. Sally has identified each of the coins with the letters A--L. Information on a weighing will be given by two strings of letters and then one of the words ``up'', ``down'', or ``even''. The first string of letters will represent the coins on the left balance; the second string, the coins on the right balance. (Sally will always place the same number of coins on the right balance as on the left balance.) The word in the third position will tell whether the right side of the balance goes up, down, or remains even.

输出

For each case, the output will identify the counterfeit coin by its letter and tell whether it is heavy or light. The solution will always be uniquely determined.

样例输入

1 
ABCD EFGH even 
ABCI EFJK up 
ABIJ EFGH even 

样例输出

K is the counterfeit coin and it is light.

 

解题思路：

 

很巧妙的方法，最开始真的不知道怎么下手......

先对每一个字母赋初值，如果字母的值是100，则一定不是轻的或重的；每一次up或者down对权重非100的字母更新，最后找非100的字母中绝对值最大的字母，即为所求。

# include<stdio.h>
# include<iostream>
# include<string>
using namespace std;
int abs(int a)
{
	return a>0?a:-a;
}
struct E
{
	string str1;
	string str2;
	string equal;
}a[100];

int mark[12];
int main()
{
	int n,i,j;
	while(cin>>n)
	{
		while(n--)
		{
			for(i=0;i<12;i++)
				mark[i]=0;//初始化
			
			for(i=0;i<3;i++)
			{
				cin>>a[i].str1>>a[i].str2>>a[i].equal;
				if(a[i].equal=="even")
				{
					for(j=0;j<a[i].str1.size();j++)
					{
						mark[a[i].str1[j]-'A'+0]=100;
						mark[a[i].str2[j]-'A'+0]=100;
					}
				}
				
				else if(a[i].equal=="up")//左边重
				{
					for(j=0;j<a[i].str1.size();j++)
					{
						if(mark[a[i].str1[j]-'A'+0]!=100)
							mark[a[i].str1[j]-'A'+0]++;
                        if(mark[a[i].str2[j]-'A'+0]!=100)
							mark[a[i].str2[j]-'A'+0]--;
					}
				}
				
				else
				{
					for(j=0;j<a[i].str1.size();j++)
					{
						if(mark[a[i].str1[j]-'A'+0]!=100)
							mark[a[i].str1[j]-'A'+0]--;
						if(mark[a[i].str2[j]-'A'+0]!=100)
							mark[a[i].str2[j]-'A'+0]++;
					}
					
				}
			}
			
			//寻找非正常的绝对值的最大值
			int max=0,maxi=0;
			for(i=0;i<12;i++)
			{
				/*	printf("%d ",mark[i]);*/
				if(mark[i]==100)
					continue;
				
				if(max<=abs(mark[i]))
				{
					max=abs(mark[i]);
					maxi=i;
				}
			}
			
			if(mark[maxi]<0)
				printf("%c is the counterfeit coin and it is light.\n",'A'+maxi);
			else
				printf("%c is the counterfeit coin and it is heavy.\n",'A'+maxi);
		}
	}
	return 0;
}