E. Vanya and Balloons

E. Vanya and Balloons

time limit per test

3 seconds

memory limit per test

512 megabytes

 

Vanya plays a game of balloons on the field of size n × n, where each cell contains a balloon with one of the values 0, 1, 2 or 3. The goal is to destroy a cross, such that the product of all values of balloons in the cross is maximum possible. There are two types of crosses: normal and rotated. For example:

**o**
**o**
ooooo
**o**
**o**

or

o***o
*o*o*
**o**
*o*o*
o***o

Formally, the cross is given by three integers r, c and d, such that d ≤ r, c ≤ n - d + 1. The normal cross consists of balloons located in cells (x, y) (where x stay for the number of the row and y for the number of the column), such that |x - r|·|y - c| = 0 and |x - r| + |y - c| < d. Rotated cross consists of balloons located in cells (x, y), such that |x - r| = |y - c| and |x - r| < d.

Vanya wants to know the maximum possible product of the values of balls forming one cross. As this value can be large, output it modulo 109 + 7.

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 1000) — the number of rows and columns in the table with balloons.

The each of the following n lines contains n characters '0', '1', '2' or '3' — the description of the values in balloons.

Output

Print the maximum possible product modulo 109 + 7. Note, that you are not asked to maximize the remainder modulo 109 + 7, but to find the maximum value and print it this modulo.

Examples

input

4
1233
0213
2020
0303

output

108

input

5
00300
00300
33333
00300
00300

output

19683

input

5
00003
02030
00300
03020
30000

output

108

Note

In the first sample, the maximum product is achieved for a rotated cross with a center in the cell (3, 3) and radius 1: 2·2·3·3·3 = 108.

题解:

题意：类似炸弹人游戏，但是十字的炸弹弹道都是等长的（就是4个方向的长度一样）（可以炸十、×），然后结果是所能炸到的每个格子的乘积的最大值。

题解: 设 X =2^a + 3^b . Y = 2^c + 3^d.

　　　　那么 X 和 Y 的大小关系与 log(2)*a + log(3)*b  和 log(2)*c + log(3)*d 的大小有关,所以我们就开一个数组存下每一个格子的log值,然后维护一下一个格子8个方向可延伸的最大长度,并且计算出2个前缀和,最后计算一下即可.........话说我的代码好TM挫,要写200+行,还是某位大佬的代码短,只要86行........

　　　　果断mo一发代码...

　　　　

#include <cmath>
#include <cstdio>
#include <algorithm>

using namespace std;

const int MAXN = 1E3 + 10;
const int MOD = 1E9 + 7;

int n;
char s[MAXN][MAXN];
double p[4][MAXN][MAXN];
int a[8][MAXN][MAXN];

int main(){
    scanf("%d", &n);
    for (int i = 1; i <= n; ++i)
        scanf("%s", s[i] + 1);

    double val[4] = {0., 0., log(2.), log(3.)};
    for (int i = 1; i <= n; ++i)
        for (int j = 1; j <= n; ++j){
            s[i][j] -= '0';
            p[0][i][j] = p[0][i    ][j - 1] + val[s[i][j]];
            p[1][i][j] = p[1][i - 1][j - 1] + val[s[i][j]];
            p[2][i][j] = p[2][i - 1][j    ] + val[s[i][j]];
            p[3][i][j] = p[3][i - 1][j + 1] + val[s[i][j]];
        }
    for (int i = 1; i <= n; ++i)
        for (int j = 1; j <= n; ++j)
            if (s[i][j]){
                a[0][i][j] = a[0][i    ][j - 1] + 1;
                a[1][i][j] = a[1][i - 1][j - 1] + 1;
                a[2][i][j] = a[2][i - 1][j    ] + 1;
                a[3][i][j] = a[3][i - 1][j + 1] + 1;
            }
            else
                a[0][i][j] = a[1][i][j] = a[2][i][j] = a[3][i][j] = 0;
    for (int i = n; i >= 1; --i)
        for (int j = n; j >= 1; --j)
            if (s[i][j]){
                a[4][i][j] = a[4][i    ][j + 1] + 1;
                a[5][i][j] = a[5][i + 1][j + 1] + 1;
                a[6][i][j] = a[6][i + 1][j    ] + 1;
                a[7][i][j] = a[7][i + 1][j - 1] + 1;
            }
            else
                a[4][i][j] = a[5][i][j] = a[6][i][j] = a[7][i][j] = 0;
    double ans = -1.0;
    int x = 0, y = 0, z = 0;
    for (int i = 1; i <= n; ++i)
        for (int j = 1; j <= n; ++j){
            if (!s[i][j])
                continue;

            int l = min(min(a[0][i][j], a[2][i][j]), min(a[4][i][j], a[6][i][j]));
            double res = p[0][i][j + l - 1] - p[0][i][j - l] + p[2][i + l - 1][j] - p[2][i - l][j] - val[s[i][j]];
            if (ans < res)
                ans = res, x = i, y = j, z = l;
            l = min(min(a[1][i][j], a[3][i][j]), min(a[5][i][j], a[7][i][j]));
            res = p[1][i + l - 1][j + l - 1] - p[1][i - l][j - l] + p[3][i + l - 1][j - l + 1] - p[3][i - l][j + l] - val[s[i][j]];
            if (ans < res)
                ans = res, x = i, y = j, z = -l;
        }
    if (ans < 0.)
        puts("0");
    else if (z > 0){
        unsigned ans = s[x][y];
        for (int i = -z + 1; i < z; ++i)
            if (i){
                ans = ans * s[x][y + i] % MOD;
                ans = ans * s[x + i][y] % MOD;
            }
        printf("%u
", ans);
    }
    else{
        unsigned ans = s[x][y];
        for (int i = -z - 1; i > z; --i)
            if (i){
                ans = ans * s[x + i][y + i] % MOD;
                ans = ans * s[x + i][y - i] % MOD;
            }
        printf("%u
", ans);
    }
    return 0;
}