Problem description

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?+?1for 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
解題思路：題目的意思就是有一個5*5的矩陣，其中只有一個元素值為1，其余元素的值全部為0，從元素1這個位置移到中心點（2，2）（下標從0~4）的過程中，每次只能向上、下、左、右（其中一個方向）移動一步，求最小的移動步數。簡單模擬一下過程即可推出：設1元素的坐標為(row,col)，則移動的最小步數為abs(row-2)+abs(col-2)，簡單AC。
AC代碼： 1 #include<bits/stdc++.h> 2 using namespace std; 3 int main(){ 4 int row,col,x; 5 for(int i=0;i<5;++i){ 6 for(int j=0;j<5;++j){ 7 cin>>x; 8 if(x){row=i;col=j;} 9 } 10 } 11 cout<<abs(row-2)+abs(col-2)<<endl; 12 return 0; 13 }

轉載于:https://www.cnblogs.com/acgoto/p/9122996.html

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