CF 236 A Beautiful Matrix

66 阅读 0 评论 44 点赞

我是靠谱客的博主甜美菠萝，最近开发中收集的这篇文章主要介绍CF 236 A Beautiful Matrix，觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

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

Output

Input

Output

题意：给出一个1 ，最少移动多少步可以到达最中间；找出坐标判断一下就好了；

AC代码：

#include <bits/stdc++.h>
using namespace std ;
int main()
{
int a[6][6];
int x , y ;
memset(a,0,sizeof(a));
for(int i = 1 ; i<=5;i++)
{
for(int j = 1 ; j<=5;j++)
{
cin>>a[i][j];
if(a[i][j]==1)
{
x = i ;
//*X存行
y = j ;
//*Y存列；
}
}
}
int ans = 0 ;
if(x<=3)
ans += 3-x ;
else
ans+= x - 3 ;
if(y<=3) ans+= 3 - y ;
else
ans+= y -3
;
cout<<ans<<endl;
return 0 ;
}