TZOJ:1050: Error Correction

描述

A boolean matrix has the parity property when each row and each column has an even sum, i.e. contains an even number of bits which are set. Here's a 4 x 4 matrix which has the parity property:
1 0 1 0
0 0 0 0
1 1 1 1
0 1 0 1
The sums of the rows are 2, 0, 4 and 2. The sums of the columns are 2, 2, 2 and 2.
Your job is to write a program that reads in a matrix and checks if it has the parity property. If not, your program should check if the parity property can be established by changing only one bit. If this is not possible either, the matrix should be classified as corrupt.

输入

The input will contain one or more test cases. The first line of each test case contains one integer n (n < 100), representing the size of the matrix. On the next n lines, there will be n integers per line. No other integers than 0 and 1 will occur in the matrix. Input will be terminated by a value of 0 for n.

输出

For each matrix in the input file, print one line. If the matrix already has the parity property, print "OK". If the parity property can be established by changing one bit, print "Change bit (i,j)" where i is the row and j the column of the bit to be changed. Otherwise, print "Corrupt".

样例输入

4
1 0 1 0
0 0 0 0
1 1 1 1
0 1 0 1
4
1 0 1 0
0 0 1 0
1 1 1 1
0 1 0 1
4
1 0 1 0
0 1 1 0
1 1 1 1
0 1 0 1
0

样例输出

OK
Change bit (2,3)
Corrupt

注意输出格式

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 

using namespace std;

int main()
{
	int a[101][101];
	int p1,p2;
	int n;
	while(~scanf("%d",&n),n) // ~ == !EOF
	{
		int ans1 = 0,ans2 = 0;
		bool ans3 = true;
		int p3 = 0,p4 = 0;
		/* 正常读入 */ 
		for(int i = 0;i < n;i ++)
			for(int j = 0;j < n;j ++)
				cin >> a[i][j];
					
		for(int i = 0;i < n;i ++)
		{
			/* ans1存每行总数 */ 
			for(int j = 0;j < n;j ++)
			{
				ans1 += a[i][j];
			}
			/* 判断每行奇偶性 */ 
			if(ans1 % 2 != 0)
			{
				p1 = i; // 确定错误行数 
				ans3 = false; // 确定有错误 
				p3 ++; // 统计错误数量 
			}
			
			ans1 = 0; // 清零	
		}
		
		/* 同理对每列进行处理 */ 
		for(int j = 0;j < n;j ++)
		{
			for(int i = 0;i < n;i ++)
			{
				ans2 += a[i][j];
			}
			if(ans2 % 2 != 0)
			{
				p2 = j;
				ans3 = false;
				p4 ++;
			}
			ans2 = 0;	
		}	
		/* 判断是否有错误 */ 
		if(ans3 == true)
			cout << "OK" << endl;
		else // 判断错误个数 
		{
			if(p3 + p4 != 2 || p3 == 0 || p4 == 0)
				cout << "Corrupt" << endl;
			else
				cout << "Change bit (" << p1 + 1 << "," << p2 + 1 << ")" << endl;
		}
	} 
	return 0;
}

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