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

In the land of ACM ruled a great King who became obsessed with order. The kingdom had a rectangular form, and the King divided the territory into a grid of small rectangular counties. Before dying, the King distributed the counties among his sons.

However, he was unaware that his children had developed a strange rivalry: the first heir hated the second heir, but not the rest; the second heir hated the third heir, but not the rest, and so on ...Finally, the last heir hated the first heir, but not the other heirs.

As soon as the King died, the strange rivalry among the King's sons sparked off a generalized war in the kingdom. Attacks only took place between pairs of adjacent counties (adjacent counties are those that share one vertical or horizontal border). A county X attacked an adjacent county Y whenever the owner of X hated the owner of Y. The attacked county was always conquered by the attacking brother. By a rule of honor all the attacks were carried out simultaneously, and a set of simultaneous attacks was called a battle. After a certain number of battles, the surviving sons made a truce and never battled again.

For example, if the King had three sons, named 0, 1 and 2, the figure below shows what happens in the first battle for a given initial land distribution:

You were hired to help an ACM historian determining, given the number of heirs, the initial land distribution and the number of battles, what was the land distribution after all battles.

Input 

The input contains several test cases. The first line of a test case contains four integers N, R, C and K, separated by single spaces. N is the number of heirs (2N100), R and C are the dimensions of the kingdom (2R, C100), and K is the number of battles (1K100). Heirs are identified by sequential integers starting from zero (0 is the first heir, 1 is the second heir, ..., N - 1 is the last heir). Each of the next R lines contains C integers H_{r, c} separated by single spaces, representing initial land distribution: H_{r, c} is the initial owner of the county in row r and column c (0H_{r, c}N - 1).

The last test case is followed by a line containing four zeroes separated by single spaces.

Output 

For each test case, your program must print R lines with C integers each, separated by single spaces in the same format as the input, representing the land distribution after all battles.

Sample Input 

3 4 4 3
0 1 2 0
1 0 2 0
0 1 2 0
0 1 2 2
4 2 3 4
1 0 3 
2 1 2 
8 4 2 1
0 7 
1 6 
2 5 
3 4 
0 0 0 0

Sample Output 

2 2 2 0
2 1 0 1
2 2 2 0
0 2 0 0
1 0 3 
2 1 2 
7 6 
0 5 
1 4 
2 3

---

兄弟們互相仇視，關係如一個環狀，每隔一個時間點勢力會被占據，問多少時間後的勢力情況。

#include <stdio.h>
#include <string.h>

int main() {
    int N, R, C, K;
    while(scanf("%d %d %d %d", &N, &R, &C, &K) == 4) {
        if(N+R+C+K == 0)
            break;
        int g[2][105][105];
        int i, j, k;
        for(i = 0; i < R; i++)
            for(j = 0; j < C; j++)
                scanf("%d", &g[0][i][j]);
        int flag = 1;
        int dx[] = {0,0,1,-1};
        int dy[] = {1,-1,0,0};
        while(K--) {
            for(i = 0; i < R; i++) {
                for(j = 0; j < C; j++) {
                    g[flag][i][j] = g[!flag][i][j];
                    int x = g[!flag][i][j]-1;
                    if(x < 0)   x += N;
                    for(k = 0; k < 4; k++) {
                        int tx = i+dx[k], ty = j+dy[k];
                        if(tx < 0 || ty < 0 || tx >= R || ty >= C)
                            continue;
                        if(g[!flag][tx][ty] == x) {
                            g[flag][i][j] = x;
                            break;
                        }
                    }
                }
            }
            flag = 1-flag;
        }
        flag = 1-flag;
        for(i = 0; i < R; i++, puts("")) {
            for(j = 0; j < C; j++) {
                if(j)   putchar(' ');
                printf("%d", g[flag][i][j]);
            }
        }
    }
    return 0;
}