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+requests+unittest执行接口自动化测试详情
查看>>
Python+requests+unittest执行接口自动化测试详情
查看>>
python+requests+unittest执行自动化接口测试!
查看>>
Python+Requests编码识别Bug
查看>>
python函数编写_[零基础学python]传说中的函数编写条规
查看>>
Python+selenium —— UI自动化之鼠标操作
查看>>
python函数注释,参数后面加冒号:,函数后面的箭头→是什么?
查看>>
Python+Selenium+Threading进行兼容性测试
查看>>
Python+selenium+unittest的GUI自动化框架实现
查看>>
python函数拟合不规则曲线_Python计算&绘图——曲线拟合问题(转)
查看>>
python函数定义的基本格式_零基础学python-2.19 定义函数、调用函数与默认参数
查看>>
Python+Selenium之数据驱动测试的实现
查看>>
python函数定义与使用+返回值简解
查看>>
Python+Selenium登录
查看>>
Python日期时间模块
查看>>
Python函数合集:足足68个内置函数请收好!
查看>>
Python+Selenium自动化测试项目实战
查看>>
Python+Selenium自动化测试项目实战【建议收藏】
查看>>
Python+Selenium自动化测试:Page Object模式
查看>>
python+selenium进行cnblog的自动化登录测试
查看>>