CodeForces刷题之路---263A - Beautiful Matrix_#p263a. beautiful matrix c++

题目：

A. Beautiful Matrix

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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:

1. Swap two neighboring matrix rows, that is, rows with indexes i and i + 1 for some integer i (1 ≤ i < 5).
2. 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

 

Examples

input

 

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

 

3

input

 

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

1

思路：

5*5的矩阵  当1处在 3行 3列 时被称为漂亮矩阵；

只需要求 1的位置与 3行 3列 位置的绝对值就可以了

用到的知识  cmath 库里的abs 函数  求绝对值

代码：

#include<iostream>
#include<cmath>
using namespace std;
 
int main()
{
    int m[6][6];
    int x , y;//用来记录1的位置
    int i,j;
    int k;
    for( i = 1 ; i < 6 ; i++ )
        for( j = 1 ; j < 6 ; j++ )
        {
            cin >> m[i][j];
            if ( m[i][j] == 1)
            {
                x = i ;
                y = j ;
            }
        }
    k = abs( x - 3 ) + abs( y - 3);
    cout << k <<endl;
    return 0;
}