Timus 1792. Hamming Code 题解

Input

Output

Samples

0 1 0 1 1 0 1

0 1 0 0 1 0 1

1 1 1 1 1 1 1

1 1 1 1 1 1 1

这样的题目就是考理解题目的能力了。

因为数据十分小，所以使用上面方法都不会超时的，只要答案对就可以了。

所以遇上这样的题目就暴力吧。

#include <iostream>
using namespace std;

void countHammingCode(int B[])
{
	B[4] = (B[1] + B[2] + B[3])%2;
	B[5] = (B[0] + B[2] + B[3])%2;
	B[6] = (B[0] + B[1] + B[3])%2;
}

void HammingCode()
{
	int A[7], B[7];
	for (int i = 0; i < 7; i++)
	{
		cin>>A[i];
		B[i] = A[i];
	}
	bool ok = false;
	for (int k = -1; k < 4; k++)
	{
		if (-1 != k) B[k] = !B[k];
		countHammingCode(B);
		ok = true;
		for (int i = 4; i < 7; i++)
		{
			if (A[i] != B[i]) ok = false;
		}
		if (ok) 
		{
			for (int i = 0; i < 7; i++)
			{
				cout<<B[i]<<‘ ‘;
			}
			break;
		}
		if (-1 != k) B[k] = !B[k];
	}
	if (!ok)
	{
		countHammingCode(B);
		for (int i = 4; i < 7; i++)
		{
			A[i] = !A[i];
			ok = true;
			for (int j = 4; j < 7; j++)
			{
				if (A[j] != B[j]) ok = false;
			}
			if (ok) 
			{
				for (int j = 0; j < 7; j++)
				{
					cout<<B[j]<<‘ ‘;
				}
				break;
			}
			A[i] = !A[i];
		}
	}
}

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

觉得上面代码不够简洁的，可以参考下面代码：

void mainHammingCode()
{
	int i,m[8];
	for(i=1; i<8; i++) scanf("%d", &m[i]);

	for(i=1; i<9; i++)
	{
		if( (m[2]+m[3]+m[4])%2 == m[5] &&
			(m[1]+m[3]+m[4])%2 == m[6] &&
			(m[1]+m[2]+m[4])%2 == m[7]) break;
		m[i]=(!m[i]);
		m[i-1]=(!m[i-1]); 
	}
	for(i=1;i<8;i++) printf("%d ",m[i]);
}

Timus 1792. Hamming Code 题解,布布扣,bubuko.com

Timus 1792. Hamming Code 题解

原文：http://blog.csdn.net/kenden23/article/details/24535723