晕
找老乡吗这是......
帮忙瞅一眼啊

的确疏忽题目的本来意思.
貌似做排列也要N久才可以出答案.
这个级别是N!*N!的.
首先一行中N个数有N!种排法,其次再来N行做排列.

好像很不容易的样子啊

main()
{int i,j;
for(i=1;i<6;i++)
{for(j=i;j<6;j++)
printf("%3d",j);
for(j=4;j>5-i;j--)
printf("%3d",j);
printf("\n");}}

悟空。。。看看！

:-)一会再写

发重复了。。。SORRY

[此贴子已经被作者于2007-6-24 18:39:43编辑过]

LINUX下编译，输出:
[JESSE@localhost tmp]$ ./a.out
N = 10
1 2 3 4 5 6 7 8 9 10
2 3 4 5 6 7 8 9 10 1
3 4 5 6 7 8 9 10 1 2
4 5 6 7 8 9 10 1 2 3
5 6 7 8 9 10 1 2 3 4
6 7 8 9 10 1 2 3 4 5
7 8 9 10 1 2 3 4 5 6
8 9 10 1 2 3 4 5 6 7
9 10 1 2 3 4 5 6 7 8
10 1 2 3 4 5 6 7 8 9

/**************************************************************/
/* phalanx.c:
* 输出N阶拉丁方阵.
*
* <June 25, 2007> JESSE
*/

#include <stdio.h>

void phalanx(unsigned n, unsigned p[][n])
{
unsigned i, j;
unsigned num, rnum;

for (i=0, num=rnum=1; i < n; i++) {
for (j=0; j < n; j++, num++) {
p[i][j] = num;
if (num == n)
num = 0;
}
num = ++rnum;
}
}
/**************************************************************/
int main(void)
{
unsigned i, j, N;

printf("N = ");
scanf("%u", &N);
/*
* init...
*/
unsigned P[N][N];
for (i=0; i < N; i++)
for (j=0; j < N; j++)
P[i][j] = 0;
/*
* calculating...
*/
phalanx(N, P);

/*
* Print them!!!
*/
for (i=0; i < N; i++) {
for (j=0; j < N; j++)
printf("%u ", P[i][j]);
printf("\n");
}

return 0;
}

[此贴子已经被作者于2007-6-24 18:41:34编辑过]