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
MySQL底层概述—4.InnoDB数据文件
查看>>
MySQL底层概述—5.InnoDB参数优化
查看>>
MySQL底层概述—6.索引原理
查看>>
MySQL底层概述—7.优化原则及慢查询
查看>>
MySQL底层概述—8.JOIN排序索引优化
查看>>
MySQL底层概述—9.ACID与事务
查看>>
Mysql建立中英文全文索引(mysql5.7以上)
查看>>
mysql建立索引的几大原则
查看>>
Mysql建表中的 “FEDERATED 引擎连接失败 - Server Name Doesn‘t Exist“ 解决方法
查看>>
MySQL开源工具推荐,有了它我卸了珍藏多年Nactive!
查看>>
MySQL异步操作在C++中的应用
查看>>
MySQL引擎讲解
查看>>
Mysql当前列的值等于上一行的值累加前一列的值
查看>>
MySQL当查询的时候有多个结果,但需要返回一条的情况用GROUP_CONCAT拼接
查看>>
MySQL必知必会(组合Where子句,Not和In操作符)
查看>>
MySQL必知必会总结笔记
查看>>
MySQL快速入门
查看>>
MySQL快速入门——库的操作
查看>>
mysql快速复制一张表的内容,并添加新内容到另一张表中
查看>>
mysql快速查询表的结构和注释,字段等信息
查看>>