Codeforces刷题之路——263A Beautiful Matrix

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我是靠谱客的博主笑点低睫毛膏，这篇文章主要介绍Codeforces刷题之路——263A Beautiful Matrix，现在分享给大家，希望可以做个参考。

You've got a 5 × 5 matrix, consisting of 24 zeroes and a single number one. Let's index the matrix rows by numbers from 1 to 5 from top to bottom, let's index the matrix columns by numbers from 1 to 5 from left to right. In one move, you are allowed to apply one of the two following transformations to the matrix:

Swap two neighboring matrix rows, that is, rows with indexes i and i + 1 for some integer i (1 ≤ i < 5).
Swap two neighboring matrix columns, that is, columns with indexes j and j + 1 for some integer j (1 ≤ j < 5).

You think that a matrix looks beautiful, if the single number one of the matrix is located in its middle (in the cell that is on the intersection of the third row and the third column). Count the minimum number of moves needed to make the matrix beautiful.

Input

The input consists of five lines, each line contains five integers: the j-th integer in the i-th line of the input represents the element of the matrix that is located on the intersection of the i-th row and the j-th column. It is guaranteed that the matrix consists of 24 zeroes and a single number one.

Output

Print a single integer — the minimum number of moves needed to make the matrix beautiful.

Examples

input
复制代码0 0 0 0 0
0 0 0 0 1
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 1
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0

output
复制代码3
3

input
复制代码0 0 0 0 0
0 0 0 0 0
0 1 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 1 0 0 0
0 0 0 0 0
0 0 0 0 0

output
复制代码11

解题思路：这是一道水题，只需记录数字“1”所在的二维坐标，进行与坐标(3,3)的之差绝对值计算即可。

提醒：注意循环一开始，所定义的 i=0 还是 i=1

以下是解题源代码（java实现）

复制代码

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import java.util.Scanner;
public class Main {
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner scanner = new Scanner(System.in);
int[][] number = new int[5][5];
int r = 0;
int l = 0;
for(int i = 0;i < 5;i++){
for (int j = 0; j < 5; j++) {
number[i][j] = scanner.nextInt();
if(number[i][j] == 1){
r = i;
l = j;
}
}
}
System.out.println(Math.abs(r-2) + Math.abs(l-2));
}
}