A. Beautiful Matrix

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.

A. Beautiful Matrix_第1张图片

题意很简单,就是看1所在的位置到中心要走多少步,一步只能走上下左右一个方向,且只有一格!

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
typedef long long ll;
#define INF 0x3f3f3f3f
#define mem memset
#define sc scanf
#define pr printf
using namespace std;
inline int read(){
	int X = 0, w = 0; char ch = 0;
	while (!isdigit(ch)) { w |= ch == '-'; ch = getchar(); }
	while (isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();
	return w ? -X : X;
}
//inline double dbread(){
//	double X = 0, Y = 1.0; int w = 0; char ch = 0;
//	while (!isdigit(ch)) { w |= ch == '-'; ch = getchar(); }
//	while (isdigit(ch)) X = X * 10 + (ch ^ 48), ch = getchar();
//	ch = getchar();//读入小数点
//	while (isdigit(ch)) X += (Y /= 10) * (ch ^ 48), ch = getchar();
//	return w ? -X : X;
//}
inline void write(int x){
	if (x < 0) putchar('-'), x = -x;
	if (x > 9) write(x / 10);
	putchar(x % 10 + '0');
}
int main() {
	ios::sync_with_stdio(false);
	int l, r, x;
	for (int i = 0; i < 5; ++i){
		for (int j = 0; j < 5; ++j) {
			x = read();
			if (x) { l = i; r = j; }
		}
	}
	write(abs(l - 2) + abs(r - 2));
	cout << endl;
	return 0;
}

 

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