学步园

Description

Input

Output

Sample input 
12 
ABCD EFGH even 
ABCI EFJK up 
ABIJ EFGH even 
AGHL BDEC even 
JKI ADE up 
J K even 
ABCDEF GHIJKL up 
ABC DEF even 
I J down 
ABCDEF GHIJKL up 
ABHLEF GDIJKC down 
CD HA even 
A B up 
B A down 
A C even 
A B up 
B C even 
DEFG HIJL even 
ABC DEJ down 
ACH IEF down 
AHK IDJ down 
ABCD EFGH even 
AB IJ even 
A L down 
EFA BGH down 
EFC GHD even 
BA EF down 
A B up 
A C up 
L K even 
ACEGIK BDFHJL up 
ACEGIL BDFHJK down 
ACEGLK BDFHJI down 
ACEGIK BDFHJL up 
ACEGIL BDFHJK down 
ACEGLK BDFHJI up 

Sample output 
K is the counterfeit coin and it is light. 
I is the counterfeit coin and it is heavy. 
I is the counterfeit coin and it is light. 
L is the counterfeit coin and it is light. 
B is the counterfeit coin and it is light. 
A is the counterfeit coin and it is heavy. 
A is the counterfeit coin and it is light. 
L is the counterfeit coin and it is heavy. 
A is the counterfeit coin and it is light. 
A is the counterfeit coin and it is heavy. 
L is the counterfeit coin and it is light. 
K is the counterfeit coin and it is heavy.

一道有意思的模拟题！

思路：使用数组标记的方法，初始化为0，重的++，轻的--, 数组里面绝对值最大的即为答案，因为怀疑最多次！

#include<iostream>
#include<cstring>
#include<cmath>

using namespace std;

const int silver = 2147483647;
int maximum = -2147483647;
int position;
int main()
{
	int time = 3;
	int coin[12];
	char status[5];
	int test;
	cin >> test;
	while(--test+1)
	{
		memset(coin,0,sizeof(coin));
		maximum = -2147483647;
		time = 3;
		while(--time+1)
		{
			char c1[7],c2[7];
			cin >> c1 >> c2 >> status;
			
			if(status[0] == 'e')
			{
				for(int i = 0;i < strlen(c1); ++i)
				{
				coin[c1[i] - 'A'] = silver;
				coin[c2[i] - 'A'] = silver;
				}
			}
			if(status[0] == 'u')
			{
				for(int i = 0; i < strlen(c1); ++i)
				{
					if(coin[c1[i] - 'A'] != silver)
						coin[c1[i] - 'A'] += 1;
					if(coin[c2[i] - 'A'] != silver)
						coin[c2[i] - 'A'] -= 1;
				}
			}
			if(status[0] == 'd')
			{
				for(int i = 0; i < strlen(c1); ++i)
				{
					if(coin[c1[i] - 'A'] != silver)
						coin[c1[i] - 'A'] -= 1;
					if(coin[c2[i] - 'A'] != silver)
						coin[c2[i] - 'A'] += 1;
				}
			}
		}
			for(int i = 0;i < 12;++i)
			{
				if(coin[i] == silver) continue;
				if(abs(coin[i]) > maximum)
				{
					maximum = abs(coin[i]);
					position = i;
				}
			}
			if(coin[position] < 0)
				cout << static_cast<char>( position + 'A') << " is the counterfeit coin and it is light."<<endl;
			if(coin[position] > 0)
				cout << static_cast<char>( position + 'A') << " is the counterfeit coin and it is heavy."<<endl;
	}
	return 0;
}