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
POJ2891:Strange Way to Express Integers——题解
查看>>
poj3045 Cow Acrobats(二分最大化最小值)
查看>>
poj3061 Subsequence(尺取法)
查看>>
poj3074 DLX精确覆盖
查看>>
poj3252(组合数)
查看>>
Qt笔记——QToolBox开发简易QQ聊天界面
查看>>
poj3307
查看>>
Qt笔记——QString与隐式共享、MVC架构
查看>>
Qt笔记——QSemaphore处理生产者/消费者模式
查看>>
Qt笔记——QMutex&QWaitCondition处理生产者消费者模式
查看>>
Qt笔记——QLable+QPixmap图片缩放踩坑
查看>>
Qt笔记——foreach与forever
查看>>
QT程序怎么挪到Linux下,linux+Qt程序如何打包发布
查看>>
Qt知识:视图框架QGraphicsWidget详解
查看>>
SpringBoot中项目启动及定时任务缓存数据库常用数据至内存变量并转换后高频调用
查看>>
Qt知识: 画刷风格
查看>>
QT的OpenGL渲染窗QOpenGLWidget Class
查看>>
QT的C++程序加载动态链接库DLL(Linux下是so)的方式
查看>>
QT界面操作1:如何跟踪鼠标位置?
查看>>
Qt环境搭建(Visual Studio)
查看>>