求一个C语言程序设计：螺旋矩阵． 求C语言程序，N*N的螺旋矩阵。
\u6c42\u4e00\u4efdC\u8bed\u8a00\u505a\u7684\u6570\u5b57\u87ba\u65cb\u77e9\u9635\u7684\u7a0b\u5e8f#include
#define N 6
int main(void)
{
int i=0,j=0,m=0,n=0,num=0;
int a[N][N];
do
{
for(i=m;i<N-m;i++)
a[i][j]=++num;
--i;
if(num==N*N)break;
for(j=m+1;j<N-m;j++)
a[i][j]=++num;
j--;
if(num==N*N)break;
for(i=N-m-2;i>=m;i--)
a[i][j]=++num;
i++;
if(num==N*N)break;
for(j=N-m-2;j>m;j--)
a[i][j]=++num;
j++;
if(num==N*N)break;
++m;
}while(num<N*N);
printf("\t\t%d\u884c%d\u5217\u87ba\u65cb\u77e9\u9635\n\n",N,N);
for(i=0;i<N;i++)
{
for(j=0;j<N;j++)
printf("%6d",a[i][j]);
printf("\n");
}
}
#include
#include
void main()
{
int i=0,j=0,n=0;
printf("\u8bf7\u8f93\u5165n:");
scanf("%d",&n);//\u77e9\u9635\u9636\u6570
int **p=NULL;//\u4e8c\u7ef4\u6307\u9488\uff0c\u5b58\u653e\u77e9\u9635n*n\u4e2a\u5143\u7d20
p= (int**)malloc(n*sizeof(int*));//\u5148\u5206\u914dn\u4e2a\u4e00\u7ef4\u6307\u9488
if(NULL==p)
exit(1);
for (i = 0; i < n ; i++)
{
p[i] = (int*)malloc(n*sizeof(int));//\u4e3an\u4e2a\u4e00\u7ef4\u6307\u9488\u5206\u914dn\u4e2a\u6574\u578b\u5927\u5c0f\u7684\u7a7a\u95f4
if (NULL==p[i])
exit(1);
}
int flag=n;//\u5269\u4e0b\u77e9\u9635\u7684\u9636\u6570\u63a7\u5236\u5faa\u73af
int row=0,cloumn=0;//\u884c\u5217\u6807\u8bb0\uff0c\u521d\u59cb\u8ba1\u6570\u4e3a0
while(flag)
{
if (flag%2)//\u77e9\u9635\u9636\u6570\u4e3a\u5947\u6570\uff0c\u4e3a\u5de6\u4e0b\u4fa7\u884c\u5217\u8d4b\u503c
{
for (i=0;i<flag;i++)
p[row+flag-1][cloumn+flag-1-i]=flag*flag-i;
for (j=1;j<flag;j++)
p[row+flag-1-j][cloumn]=p[row+flag-1][cloumn]-j;
cloumn++;//\u62c6\u53bb\u4e00\u884c\u4e00\u5217\uff0c\u5f71\u54cd\u4e0b\u6b21\u5faa\u73af\u5217\u4e0b\u6807\uff0c\u52a01
}
else//\u4e3a\u5076\u6570\uff0c\u4e3a\u53f3\u4e0a\u4fa7\u884c\u5217\u8d4b\u503c
{
for (i=0;i<flag;i++)
p[row][cloumn+i]=flag*flag-i;
for (j=1;j<flag;j++)
p[row+j][cloumn+flag-1]=p[row][cloumn+flag-1]-j;
row++;//\u62c6\u53bb\u4e00\u884c\u4e00\u5217\uff0c\u5f71\u54cd\u4e0b\u6b21\u5faa\u73af\u884c\u4e0b\u6807\uff0c\u52a01
}
flag--;//\u5269\u4e0b\u77e9\u9635\u7684\u9636\u6570
}
printf("%d*%d\u9636\u77e9\u9635\u4e3a\uff1a\n",n,n);
i=n;
while(i--)
printf("********");
printf("\n");
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
printf("%d\t",p[i][j]);
printf("\n");
}
for (i = 0; i < n; i++)
free(p[i]);
free(p);
}
#include<stdio.h>
main()
{
int a,b,k,n,i,j,a1,b1,m;
int lines[50][50];
scanf("%d",&n);
i=j=0;a=b=n;a1=b1=0;k=1;
if(n==1)
printf(" 1\n");
else
{
for(m=1;m<2*n;m++)
{
if(m%4==1)
{
for(;j<b;j++)
{
lines[i][j]=k;
k++;
}
i++;
j--;
b--;
}
else if(m%4==2)
{
for(;i<a;i++)
{
lines[i][j]=k;
k++;
}
j--;
i--;
a--;
}
else if(m%4==3)
{
for(;j>=b1;j--)
{
lines[i][j]=k;
k++;
}
i--;
j++;
b1++;
}
else
{
for(;i>=a1+1;i--)
{
lines[i][j]=k;
k++;
}
j++;
i++;
a1++;
}
}
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{ if(j==0)printf(" %2d ",lines[i][j]);
else if(j==n-1)printf("%2d",lines[i][j]);
else
printf("%2d ",lines[i][j]);
}
printf("\n");
}
}
}
和的话加一下就是了；
至于你说的那些个积什么的，比较麻烦。自己再弄弄吧。
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