B Error correction

最新推荐文章于 2022-11-19 15:46:45 发布

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本文介绍了一种算法，用于检查给定矩阵是否满足行和列的元素总和均为偶数的条件，即所谓的奇偶校验属性。如果矩阵不满足此属性，算法还会尝试通过更改单个元素来实现这一属性。

题目描述

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:

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.

Input

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.

Output

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".

Sample Input

Sample Output

OK
Change bit (2,3)
Corrupt

题目原网站：http://poj.org/problem?id=2260

思路：分3种情况

1.行和或列和中有两个或两个以上为奇数，则输出corrupt。

2.行和与列和均为偶数，输出OK。

3. 行和与列和均只有一个为奇数，则只需改动该行与该列所在的数即可。

代码如下：

 #include<iostream>
#include<stdio.h>
using namespace std;
int main(void) {
	int n;
	while (cin >> n&&n) {
		int *sumc = new int[n], *sumr = new int[n], c, r, oddc = 0, oddr = 0;
		for(int i=0;i<n;i++)
			sumc[i]=0,sumr[i]=0;
		int **a = new int*[n];
		for (int i = 0; i < n; i++)
			a[i] = new int[n];
		for (int i = 0; i < n; i++)
			for (int j = 0; j < n; j++)
			{
				cin >> a[i][j];
				sumr[i] += a[i][j];
			}
		for (int j = 0; j < n; j++)
			for (int i = 0; i < n; i++)
				sumc[j] += a[i][j];
		for (int i = 0; i < n; i++)
		{
			if (sumr[i] & 1) oddr++, r = i;
			if (sumc[i] & 1) oddc++, c = i;
		}
		if (oddr > 1 || oddc > 1) cout << "Corrupt" << endl;
		if (oddr == 1 && oddc == 1) printf("Change bit (%d,%d)\n",r+1,c+1);
		if (oddr == 0 && oddc == 0) printf("OK\n");
	}
	return 0;
}