codeforces711B Chris and Magic Square 简单枚举

B. Chris and Magic Square

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

ZS the Coder and Chris the Baboon arrived at the entrance of Udayland. There is a n × n magic grid on the entrance which is filled with integers. Chris noticed that exactly one of the cells in the grid is empty, and to enter Udayland, they need to fill a positive integer into the empty cell.

Chris tried filling in random numbers but it didn't work. ZS the Coder realizes that they need to fill in a positive integer such that the numbers in the grid form a magic square. This means that he has to fill in a positive integer so that the sum of the numbers in each row of the grid (), each column of the grid (), and the two long diagonals of the grid (the main diagonal —  and the secondary diagonal — ) are equal.

Chris doesn't know what number to fill in. Can you help Chris find the correct positive integer to fill in or determine that it is impossible?

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 500) — the number of rows and columns of the magic grid.

n lines follow, each of them contains n integers. The j-th number in the i-th of them denotes ai, j (1 ≤ ai, j ≤ 109 or ai, j = 0), the number in the i-th row and j-th column of the magic grid. If the corresponding cell is empty, ai, j will be equal to 0. Otherwise, ai, j is positive.

It is guaranteed that there is exactly one pair of integers i, j (1 ≤ i, j ≤ n) such that ai, j = 0.

Output

Output a single integer, the positive integer x (1 ≤ x ≤ 1018) that should be filled in the empty cell so that the whole grid becomes a magic square. If such positive integer x does not exist, output  - 1 instead.

If there are multiple solutions, you may print any of them.

Examples

input

3
4 0 2
3 5 7
8 1 6

output

9

input

4
1 1 1 1
1 1 0 1
1 1 1 1
1 1 1 1

output

1

input

4
1 1 1 1
1 1 0 1
1 1 2 1
1 1 1 1

output

-1

Note

In the first sample case, we can fill in 9 into the empty cell to make the resulting grid a magic square. Indeed,

The sum of numbers in each row is:

4 + 9 + 2 = 3 + 5 + 7 = 8 + 1 + 6 = 15.

The sum of numbers in each column is:

4 + 3 + 8 = 9 + 5 + 1 = 2 + 7 + 6 = 15.

The sum of numbers in the two diagonals is:

4 + 5 + 6 = 2 + 5 + 8 = 15.

In the third sample case, it is impossible to fill a number in the empty square such that the resulting grid is a magic square.

傻逼了，不知道一开始怎么想的……居然还你妹 Wrong answer on test 116 

还WA on 116了好几发……

幸亏调过了……那尼玛也很脑残啊……当时怎么想的……

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cstdlib>
#include <iostream>
#include <set>

using namespace std;

typedef long long ll;

const int maxn = 500;
int n;
ll a[maxn + 5][maxn + 5], res, ans, ans2;
bool flag, f, f2;

int main()
{
	ios::sync_with_stdio(false);
	cin >> n;
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= n; j++) {
			cin >> a[i][j];
		}
	}
	if (n == 1) {
		cout << 1 << endl;
		return 0;
	}
	for (int i = 1; i <= n; i++) {
		flag = true; res = 0;
		for (int j = 1; j <= n; j++) {
			res += a[i][j];
			if (a[i][j] == 0) {
				flag = false;
				break;
			}
		}
		if (flag) break;
	}
#if 0
	cout << "res: " << res << endl;
#endif
	set<ll> s;
	for (int i = 1; i <= n; i++) {
		ans = 0; f = false;
		for (int j = 1; j <= n; j++) {
			ans += a[i][j];
			if (a[i][j] == 0) f = true;
		}
		if (f && ans >= res) {
			cout << -1 << endl;
			return 0;
		}
		if (res != ans) {
			if (f) {
				s.insert(res - ans);
			}
			else {
				puts("-1");
				return 0;
			}
		}
	}
#if 0
	cout << "ok1" << endl;
	cout << s.size() << endl;
	cout << *(s.begin()) << endl;
#endif
	for (int i = 1; i <= n; i++) {
		ans = 0; f = false;
		for (int j = 1; j <= n; j++) {
			ans += a[j][i];
			if (a[j][i] == 0) f = true;
		}
		if (f && ans >= res) {
			cout << -1 << endl;
			return 0;
		}
		if (res != ans) {
			if (f)
				s.insert(res - ans);
			else {
				puts("-1");
				return 0;
			}
		}
	}
#if 0
	cout << "ok2" << endl;
	cout << s.size() << endl;
	cout << *(s.begin()) << endl;
#endif
	ans = ans2 = 0; f = f2 = false;
	for (int i = 1; i <= n; i++) {
		ans += a[i][i];
		ans2 += a[i][n - i + 1];
		if (a[i][i] == 0) f = true;
		if (a[i][n - i + 1] == 0) f2 = true;
	}
#if 0
	cout << "testok" << endl;
	cout << ans << endl;
	cout << ans2 << endl;
#endif
	if ((f && ans >= res) || (f2 && ans2 >= res)) {
		cout << -1 << endl;
		return 0;
	}
#if 0
	cout << "ok34" << endl;
	cout << s.size() << endl;
	cout << *(s.begin()) << endl;
#endif
	if (res != ans) {
		if (f) {
			s.insert(res - ans);
		}
		else {
			puts("-1");
			return 0;
		}
	}
	if (res != ans2) {
		if (f2) {
			s.insert(res - ans2);
		}
		else {
			puts("-1");
			return 0;
		}
	}
	if (s.size() == 1 && *(s.begin()) >= 1 && *(s.begin()) <= 1e18) {
		cout << *(s.begin()) << endl;
	}
	else {
		cout << -1 << endl;
	}
	return 0;
}

转念一想，你妹判断个毛线啊，把该填的数直接填进去然后直接扫判断不就行了么，一发AC，扇自己一巴掌……啪……

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cstdlib>
#include <iostream>
#include <set>

using namespace std;

typedef long long ll;

const int maxn = 500;
int n;
ll a[maxn + 5][maxn + 5], res, ans, x, y, tmp, tmp1, tmp2;
bool flag;

int main()
{
	ios::sync_with_stdio(false);
	cin >> n;
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= n; j++) {
			cin >> a[i][j];
			if (a[i][j] == 0) {
				x = i; y = j;
			}
		}
	}
	if (n == 1) {
		cout << 1 << endl;
		return 0;
	}
	for (int i = 1; i <= n; i++) {
		flag = true; res = 0;
		for (int j = 1; j <= n; j++) {
			res += a[i][j];
			if (a[i][j] == 0) {
				flag = false;
				break;
			}
		}
		if (flag) break;
	}
	ans = 0;
	for (int j = 1; j <= n; j++) {
		ans += a[x][j];
	}
	if (ans < res) {
		a[x][y] = res - ans;
		flag = true;
		for (int i = 1; i <= n; i++) {
			tmp1 = tmp2 = 0;
			for (int j = 1; j <= n; j++) {
				tmp1 += a[i][j];
				tmp2 += a[j][i];
			}
			if (tmp1 != res || tmp2 != res) {
				flag = false;
				break;
			}
		}
		if (!flag) {
			cout << -1 << endl;
			return 0;
		}
		tmp1 = tmp2 = 0;
		for (int i = 1; i <= n; i++) {
			tmp1 += a[i][i];
			tmp2 += a[i][n - i + 1];
		}
		if (tmp1 != res || tmp2 != res) {
			cout << -1 << endl;
			return 0;
		}
		cout << a[x][y] << endl;
	}
	else {
		cout << -1 << endl;
	}
	return 0;
}