Codeforces——263A Beautiful Matrix

小声BB

又水了一天，今天开始补

题干

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
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

分析

这道题比较简单，就是算“1”的坐标，然后abs（-3）就行

代码

#include<iostream>
#include<math.h>
using namespace std;
int main()
{
int a[6][6],x,y;
for(int i = 1;i < 6;i++)
for(int j = 1;j < 6;j++)
{
cin >> a[i][j];
if(a[i][j] == 1)
{
x = i;
y = j;
}
}
cout << abs(x - 3) + abs(y - 3);
return 0;
}

最后

以上就是淡然母鸡为你收集整理的Codeforces——263A Beautiful Matrix的全部内容，希望文章能够帮你解决Codeforces——263A Beautiful Matrix所遇到的程序开发问题。

如果觉得靠谱客网站的内容还不错，欢迎将靠谱客网站推荐给程序员好友。