題目來源：http://codeforces.com/problemset/problem/263/A

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
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

題目大意：

有一個5*5的矩陣，裡面有有24個0，1個1。可以將相鄰的兩行或兩列進行交換。如果1在中心位置（第三行、第三列），我們就稱矩陣是美麗的。請問給出一個矩陣，求最少進行多少次交換可以得到美麗的矩陣。

題目分析：

模擬題，首先讀取時記錄1的位置行x，列y。最少交換次數就是x距離3的長度+y距離3的長度，只需要abs(x-3)+abs(y-3)即可。

參考程式碼：

#include <bits/stdc++.h>
using namespace std;
int main(){
	int a[6][6], x, y;
	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, y = j;
		}
	cout << abs(x-3)+abs(y-3);
	return 0;
}