POJ - Counterfeit Dollar 题解

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挺考智力的题目。

思路：

1 如果是假币，那么每次都必定引起天平的不平衡

2 如果天平平横，那么全部都肯定是真币

利用这个特性，利用hash表，就能写出很简洁的程序。

如果使用枚举，那么会（轻松？）过百行的代码的。

当然其实题目给出了条件：一定可以找出唯一的假币的。

如果没有这个条件，那么是不一定可以三次称，就能确定结果的。

下面程序参考了别人的：

http://www.cnblogs.com/orangeman/archive/2009/07/10/1520663.html

这个家伙的思路也不错，而且给出的额外的cases都是正确的：http://www.slyar.com/blog/poj-1013-c.html

原题：

http://poj.org/problem?id=1013

#include <string>
#include <vector>
#include <iostream>
using namespace std;

class CounterfeitDollar
{
	const static int ALPHA = 12;
	const static int TIMES = 3;
	string s1, s2, s3;
public:
	CounterfeitDollar()
	{
		int n;
		cin>>n;
		while (n--)
		{
			int tbl[ALPHA][TIMES] = {0};
			int balance[TIMES] = {0};
			for (int i = 0; i < TIMES; i++)
			{
				cin>>s1>>s2>>s3;
				for (int j = 0; j < (int)s1.size(); j++)
				{
					tbl[s1[j] - ‘A‘][i] = -1;
					tbl[s2[j] - ‘A‘][i] = 1;
				}
				if (‘e‘ == s3[0]) balance[i] = 0;
				else if (‘d‘ == s3[0]) balance[i] = 1;
				else balance[i] = -1;
			}
			for (int i = 0; i < ALPHA; i++)
			{
				if (balance[0] == -tbl[i][0] && balance[1] == -tbl[i][1] &&
					balance[2] == -tbl[i][2])
				{
					cout<<char(‘A‘+i)
						<<" is the counterfeit coin and it is light.\n";
					break;
				}
				else if (balance[0] == tbl[i][0] && balance[1] == tbl[i][1] 
						&& balance[2] == tbl[i][2])
				{
					cout<<char(‘A‘+i)
						<<" is the counterfeit coin and it is heavy.\n";
					break;
				}
			}
		}//while (n--)
	}
};

int counterfeitDollar()
{
	CounterfeitDollar();
	return 0;
}

更新一个新的解法：

1 如果是even，那么所有是真币，所以设置为10

2 如果硬币在轻的一方，那么--，如果在重的一方，那么++

3 最后找到差别最大的硬币，那么就为假币

4 如果假币为负数，那么就比真币轻， 如果为正，那么就比真币重。

这回是原创的程序的，感觉比前面的更加容易理解。

#include <string>
#include <vector>
#include <iostream>
using namespace std;

class CounterfeitDollar_2
{
	const static int ALPHA = 12;
	const static int TIMES = 3;
	string s1, s2, s3;
public:
	CounterfeitDollar_2()
	{
		int n;
		cin>>n;
		while (n--)
		{
			int tbl[ALPHA] = {0};
			int balance[TIMES] = {0};
			for (int i = 0; i < TIMES; i++)
			{
				cin>>s1>>s2>>s3;
				if (‘e‘ == s3[0])
				{
					for (int i = 0; i < (int)s1.size(); i++)
					{
						tbl[s1[i] - ‘A‘] = 10;
						tbl[s2[i] - ‘A‘] = 10;
					}
				}
				else if (‘d‘ == s3[0])
				{
					for (int i = 0; i < (int)s1.size(); i++)
					{
						if (tbl[s1[i] - ‘A‘] != 10) tbl[s1[i] - ‘A‘]--;
						if (tbl[s2[i] - ‘A‘] != 10) tbl[s2[i] - ‘A‘]++;
					}
				}
				else
				{
					for (int i = 0; i < (int)s1.size(); i++)
					{
						if (tbl[s1[i] - ‘A‘] != 10) tbl[s1[i] - ‘A‘]++;
						if (tbl[s2[i] - ‘A‘] != 10) tbl[s2[i] - ‘A‘]--;
					}
				}
			}
			int id = 0, diff = 0;
			for (int i = 0; i < ALPHA; i++)
			{
				if (tbl[i] != 10 && diff < abs(tbl[i]))
				{
					diff = abs(tbl[i]);
					id = i;
				}
			}
			cout<<char(‘A‘+id);
			if (tbl[id] < 0) 
				cout<<" is the counterfeit coin and it is light.\n";
			else 
				cout<<" is the counterfeit coin and it is heavy.\n";
		}//while (n--)
	}
};

int main()
{
	CounterfeitDollar_2();
	return 0;
}

POJ - Counterfeit Dollar 题解,布布扣,bubuko.com

POJ - Counterfeit Dollar 题解

标签：blog   class   code   c   http   int   

原文地址：http://blog.csdn.net/kenden23/article/details/25423381