博客
关于我
POJ 2260 Error Correction 模拟 贪心 简单题
阅读量:429 次
发布时间:2019-03-06

本文共 2135 字,大约阅读时间需要 7 分钟。

Error Correction
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 6825   Accepted: 4289

Description

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.

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

41 0 1 00 0 0 01 1 1 10 1 0 141 0 1 00 0 1 01 1 1 10 1 0 141 0 1 00 1 1 01 1 1 10 1 0 10

Sample Output

OKChange bit (2,3)Corrupt

计算每行每列1的数量,看奇数个的交叉点,大于一个或没有则不能实现。


#include<stdio.h>int main(){	int a[105][105];	int n;	while (~scanf("%d", &n))	{		int i, j;		int num1 = 0,num2=0;		if (!n)		{			break;		}		int x, y;		for ( i = 1; i <= n; i++)		{			int sum = 0;			for ( j = 1; j <= n; j++)			{				scanf("%d", &a[i][j]);				sum += a[i][j];			}			if (sum % 2 != 0)			{				x = i;				num1++;			}		}		for ( i = 1; i <= n; i++)		{			int sum = 0;			for ( j = 1; j<= n;j++)			{				sum += a[j][i];			}			if (sum % 2 != 0)			{				y = i;				num2++;			}		}				if (num1 == 0 && num2 == 0)		{			printf("OK\n");		}		else if (num1 == 1 && num2 == 1)		{			printf("Change bit (%d,%d)\n", x, y);		}		else		{			printf("Corrupt\n");		}	}	return 0;}


转载地址:http://gkwuz.baihongyu.com/

你可能感兴趣的文章
Redis五种数据结构简介
查看>>
PHPCMS多文件上传和上传数量限制
查看>>
phpEnv的PHP集成环境
查看>>
PHPExcel一些基本设置总结
查看>>
phpexcel中文手册
查看>>
PHPExcel导入导出 若在thinkPHP3.2中使用(无论实例还是静态调用(如new classname或classname::function)都必须加反斜杠,因3.2就命名空间,如/c...
查看>>
phpize及其用法
查看>>
phpMailer发送邮件
查看>>
PHPMailer发送邮件
查看>>
phpmailer发送邮件,可以带附件
查看>>
phpmailer的用法
查看>>
phpmyadmin 安装
查看>>
phpmyadmin导出数据库出现Fatal error: Cannot 'break' 2 levels in D:\phpstudy\WWW\phpMyAdmin
查看>>
phpmyadmin数据库建表及插入
查看>>
phpnow配置
查看>>
phprpc简单使用
查看>>
phpspider中当爬虫获取数据时如何去掉广告
查看>>
phpstorm 2016.3.3 激活
查看>>
phpstorm中Xdebug的使用
查看>>
phpstorm中使用svn版本控制器
查看>>