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高级-视图
查看>>
MySQL:判断逗号分隔的字符串中是否包含某个字符串
查看>>
Nacos在双击startup.cmd启动时提示:Unable to start embedded Tomcat
查看>>
Nacos安装教程(非常详细)从零基础入门到精通,看完这一篇就够了
查看>>
Nacos配置中心集群原理及源码分析
查看>>
nacos配置自动刷新源码解析
查看>>
Nacos集群搭建
查看>>
nacos集群搭建
查看>>
Navicat for MySQL 查看BLOB字段内容
查看>>
Neo4j电影关系图Cypher
查看>>
Neo4j的安装与使用
查看>>
Neo4j(2):环境搭建
查看>>
Neo私链
查看>>
nessus快速安装使用指南(非常详细)零基础入门到精通,收藏这一篇就够了
查看>>
Nessus漏洞扫描教程之配置Nessus
查看>>
Nest.js 6.0.0 正式版发布,基于 TypeScript 的 Node.js 框架
查看>>
NetApp凭借领先的混合云数据与服务把握数字化转型机遇
查看>>
NetBeans IDE8.0需要JDK1.7及以上版本
查看>>
netcat的端口转发功能的实现
查看>>
netfilter应用场景
查看>>