URAL - 1792 Hamming Code(枚举)

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Description

Let us consider four disks intersecting as in the figure. Each of the three shapes formed by the intersection of three disks will be called apetal.

Write zero or one on each of the disks. Then write on each petal the remainder in the division by two of the sum of integers on the disks that contain this petal. For example, if there were the integers 0, 1, 0, and 1 written on the disks, then the integers written on the petals will be 0, 1, and 0 (the disks and petals are given in the order shown in the figure).

This scheme is called a Hamming code. It has an interesting property: if you enemy changes secretely any of the seven integers, you can determine uniquely which integer has been changed. Solve this problem and you will know how this can be done.

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

直接枚举

#include<iostream>
using namespace std;
int a[10];
bool ok(int a1, int a2, int a3, int a4, int b1,int b2, int b3)
{
	if ((a1 + a2 + a4) % 2 != b3) return false;
	if ((a1 + a3 + a4) % 2 != b2) return false;
	if ((a2 + a3 + a4) % 2 != b1) return false;
	return true;
}
int main()
{
	while (cin>>a[0])
	{
		for (int i = 1; i < 7; i++)
			cin >> a[i];
		if (ok(a[0], a[1], a[2], a[3], a[4], a[5], a[6]))
		{
			for (int i = 0; i < 7; i++)
			{
				if (i) cout << " ";
				cout << a[i];
			}
			cout << endl;
			continue;
		}
		for (int i = 0; i < 7; i++)
		{
			a[i] = 1^a[i];
			if (ok(a[0], a[1], a[2], a[3], a[4], a[5], a[6]))
			{
				for (int i = 0; i < 7; i++)
				{
					if (i) cout << " ";
					cout << a[i];
				}
				cout << endl;
				break;
			}
			a[i] = 1^a[i];
		}	
	}
}

URAL - 1792 Hamming Code(枚举)

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原文地址：http://blog.csdn.net/qq_18738333/article/details/45157139