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
python——random模块的应用
查看>>
Python——基础知识细节
查看>>
Python——正则表达式
查看>>
Python—面向对象2
查看>>
Python“和“带整数的运算符
查看>>
Python“文件存在“制作目录时出错
查看>>
Python“文件存在“制作目录时出错
查看>>
Python、IMAP 和 GMail.将消息标记为已查看
查看>>
python一个一个获取迭代器的元素
查看>>
python一维数组 提取_python – 如何从numpy中的1维数组中提取零维切片
查看>>
python三大库之--numpy(一)
查看>>
python三大库之--numpy(二)
查看>>
python上的字无效字符_标识中的无效字符
查看>>
python下用decode和encode 中文转码
查看>>
Python下调用Linux的Shell命令
查看>>
Python接口自动化测试实战系列
查看>>
python下载_安装_配置_以及第一行python程序---python工作笔记009
查看>>
Python不协调机器人消息.内容
查看>>
Python与C#之间内存共享互传信息
查看>>
python与c语言混合编程_你试过C语言和Python一起混合编程吗?两者相加不是已经无敌了!...
查看>>