MiaoMiao
unread,Jan 13, 2008, 7:53:34 AM1/13/08Sign in to reply to author
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to 程序=数据结构+算法
#include <stdio.h>
#include <stdlib.h>
#include <assert.h>
/*
1 8 7
2 9 6
3 4 5
输出如图所示的矩正,第一个是3*3的,后面的4*4的,
算法就是随便输入一个N,就可以输出一个N*N的矩正???
1 12 11 10
2 13 16 9
3 14 15 8
4 5 6 7
*/
void MatrixProcess( int ** Matrix,int N )
{
int number=1;
int start=0,end=N,i;
assert( N > 1 );
assert( Matrix != NULL );
for(i=0; i < N/2+1; ++i)
{
int iter=0;
for( iter=start; iter<end; ++iter )
if( Matrix[iter][start] == 0 )
Matrix[iter][start] = number++;
for( iter= ++start; iter<end; ++iter )
if( Matrix[end-1][iter] == 0 )
Matrix[end-1][iter] = number++;
for( iter= --end; iter>=start-1; --iter )
if( Matrix[iter][end] == 0 )
Matrix[iter][end] = number++;
for( iter= end; iter>=start; --iter )
if( Matrix[start-1][iter] == 0 )
Matrix[start-1][iter] = number++;
}
}
int ** MatrixMemAlloc( int N )
{
/**//* N为阶数,分配N*N的二维空间,清零 */
int ** mem;
int i,j;
assert( N > 1 );
mem = (int **) malloc ( (sizeof(int *)) * N );
assert(mem != NULL);
for(i=0; i<N; ++i)
{
mem[i] = (int *) malloc ( (sizeof(int)) * N );
assert( mem[i] != NULL );
}
for(i=0; i< N; ++i)
for(j=0; j<N; ++j)
mem[i][j]=0; /* 清零 */
return mem;
}
void MatrixPrint(int ** Matrix,int N)
{
int i,j;
assert( N>1 );
assert( Matrix != NULL );
for(i=0;i<N;++i)
assert( Matrix[i] != NULL );
for(i=0;i<N;++i)
{
for(j=0;j<N;++j)
printf("%3d",Matrix[i][j]);
printf(" ");
}
return;
}
int main()
{
int N,** point;
printf("input N:");
scanf("%d",&N);
point = MatrixMemAlloc( N );
MatrixProcess( point,N );
MatrixPrint( point, N );
return 0;
}