#include<iostream>
using namespace std;void zigzag(int**M, int N)
{/** example N=5 结果*    0       1       5       6       14      15      27      28*    2       4       7       13      16      26      29      42*    3       8       12      17      25      30      41      43*    9       11      18      24      31      40      44      53*    10      19      23      32      39      45      52      54*    20      22      33      38      46      51      55      60*    21      34      37      47      50      56      59      61*    35      36      48      49      57      58      62      63**//*prestep表示上一步的动作，其值及其对应关系如下：case 1:上一步向右走case 2;上一步左下走case 3:上一步向下走case 4:上一步右上走*/int prestep;int count = 0;int i = 0;int j = 0;*(*(M + i) + j) = count;count++;prestep = 4;while (count < N*N){if (i == 0)//上边界走法
        {if (j == N - 1){if (prestep == 1)//上一步右，接着左下
                {i++; j--;prestep = 2;}else if (prestep == 4){i++;prestep = 3;}}else if (prestep == 4)//上一步右上方向走，则现在向右走一步;
            {j++;prestep = 1;}else if (prestep == 1)//上一步向右走，现在需要向左下走一步;
            {i++;j--;prestep = 2;}else{cerr << "error 1!" << endl;return;}}else if (i == N - 1)//下边界走法
        {if (prestep == 2)//上一步左下方向，则向右走
            {j++;prestep = 1;}else if (prestep == 1 || prestep == 3)//上一步向右走，则像右上走;
            {i--;j++;prestep = 4;}else{cerr << "error 2!" << endl;return;}}else if (j == 0)//左边界
        {if (prestep == 2)//上一步左下，这一步向下
            {i++;prestep = 3;}else if (prestep == 3)//上一步向下，这一步右上
            {i--;j++;prestep = 4;}else{cerr << "error 3!" << endl;return;}}else if (j == N - 1)//右边界
        {if (prestep == 4)//上一步右上，这一步向下
            {i++;prestep = 3;}else if (prestep == 3 || prestep == 1)//上一步向下，这一步左下
            {i++; j--;prestep = 2;}else{cerr << "error 4!" << endl;return;}}else{if (prestep == 2)//上一步左下，接着左下
            {i++; j--;prestep = 2;}else if (prestep == 4)//上一步右上，这一步接着右上
            {i--; j++;prestep = 4;}else{cerr << "error 5!" << endl;return;}}*(*(M + i) + j) = count++;}
}
int main()
{   int N;cout << "please input N:";cin >> N;int **M = (int**)malloc(sizeof(int *)*N);for (int i = 0; i < N; i++){*(M + i) = (int *)malloc(sizeof(int)*N);}zigzag(M, N);for (int i = 0; i < N; i++){for (int j = 0; j < N; j++){cout << *(*(M+i)+j) << "\t";}cout << endl;}for (int i = 0; i < N; i++){free(*(M + i));}free(M);
}