Home on the Range

Farmer John grazes his cows on a large, square field N (2 <= N <= 250) miles on a side (because, for some reason, his cows will only graze on precisely square land segments). Regrettably, the cows have ravaged some of the land (always in 1 mile square increments). FJ needs to map the remaining squares (at least 2x2 on a side) on which his cows can graze (in these larger squares, no 1x1 mile segments are ravaged).

Your task is to count up all the various square grazing areas within the supplied dataset and report the number of square grazing areas (of sizes >= 2x2) remaining. Of course, grazing areas may overlap for purposes of this report.

### INPUT FORMAT

 Line 1: N, the number of miles on each side of the field. Line 2..N+1: N characters with no spaces. 0 represents "ravaged for that block; 1 represents "ready to eat".

### SAMPLE INPUT (file range.in)

`6101111001111111111001111101101111001`

### OUTPUT FORMAT

Potentially several lines with the size of the square and the number of such squares that exist. Order them in ascending order from smallest to largest size.

### SAMPLE OUTPUT (file range.out)

```2 103 44 1  题意：给出一个正方形牧场，0表示没草皮，1表示有草皮。
/*LANG: CTASK: range*/#include<stdio.h>#define maxn 251int map[maxn][maxn];int ans[maxn], n;int min(int a, int b, int c){    int d = a < b ? a : b;    return d < c ? d : c;}void dp(){    int i, j, k;    for (i = n - 1; i >= 1; i--)    {        for (j = n - 1; j >= 1; j--)        {            //表示以map[i][j]为左上角的正map[i][j]边形            if (map[i][j])//当下面和右边以及右下边都是1是，正边形边长加1            {                map[i][j] = min(map[i+1][j+1], map[i+1][j], map[i][j+1]) + 1;            }            if (map[i][j] > 1)//对于边数> 2的，累加正变形个数,增加正边形的个数等于map[i][j]边形对角线上的单位正边形个数            {                for (k = 2; k <= map[i][j]; k++)                {                    ans[k]++;                }            }        }    }}int main(){    freopen("range.in", "r", stdin), freopen("range.out", "w", stdout);    int i, j;    scanf("%d", &n);    for (i = 1; i <= n; i++)    {        for (j = 1; j  <= n; j++)        {            scanf("%1d", &map[i][j]);        }    }    dp();    for (i = 2; i <= maxn; i++)    {        if (ans[i])        {            printf("%d %d\n", i, ans[i]);        }    }    fclose(stdin), fclose(stdout);    return 0;}