Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

Rotated squares

The Problem

Given a square of N rows and columns of uppercase letters, and another smaller square of n rows and columns of uppercase letters, we want to count the number of appearances in the big square of the small square in all the rotated forms.

The Input

The input will consist of a series of problems, with each problem in a series of lines. In the first line, the dimension of the two squares, N and n (with 0<n<=N), is indicated by two integer numbers separated by a space. The N lines of the first square appear in the following N lines of the input, and then the n lines of the second square appear in the following n lines. The characters in a line are one after another, without spaces. The end of the sequence of problems is indicated with a case where N=0 and n=0.

The Output

The solutions of the different problems appear in successive lines. For each problem the output consists of a line with four integers, which are the number of times each rotation of the small square appears in the big square. The first number corresponds to the number of appearances of the small square without rotations, the second to the appearances of the square rotated 90 degrees (clockwise), the third to the square rotated 180 degrees, and the fourth to the square rotated 270 degrees. 

Sample Input

4 2
ABBA
ABBB
BAAA
BABB
AB
BB
6 2
ABCDCD
BCDCBD
BACDDC
DCBDCA
DCBABD
ABCDBA
BC
CD
0 0

Sample Output

0 1 0 0
1 0 1 0

---

#include <stdio.h>
#include <string.h>
char big[500][500], small[500][500];
int find(int n, int m) {
    int ans = 0, i, j, k, l;
    for(i = 0; i < n; i++) {
        for(j = 0; j < n; j++) {
            if(i+m <= n && j+m <= n) {
                int cnt = 0;
                for(k = 0; k < m; k++) {
                    for(l = 0; l < m; l++) {
                        if(big[i+k][j+l] == small[k][l])
                            cnt++;
                    }
                }
                if(cnt == m*m)
                    ans++;
            }
        }
    }
    return ans;
}
void rotate(int n) {
    char tmp[n][n];
    int i, j;
    for(i = 0; i < n; i++) {
        for(j = 0; j < n; j++)
            tmp[i][j] = small[n-j-1][i];
    }
    for(i = 0; i < n; i++) {
        for(j = 0; j < n; j++)
           small[i][j] = tmp[i][j];
    }
}
int main() {
    int n, m, i;
    while(scanf("%d %d", &n, &m) == 2) {
        if(n == 0 && m == 0)
            break;
        for(i = 0; i < n; i++)
            scanf("%s", big[i]);
        for(i = 0; i < m; i++)
            scanf("%s", small[i]);
        for(i = 0; i < 4; i++) {
            if(i)
                putchar(' ');
            printf("%d", find(n, m));
            rotate(m);
        }
        puts("");
    }
    return 0;
}