#include <conio.h>
int main (void) {
int arr[20] = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20}
while (p-arr != 20) { /* 遍历数组,每元素值+1 */
(*p)++
printf ("%d\t", *p)
p++
}
putchar ('\n')
getch () /* 屏幕暂留 */
return 0
}
是指上这个问题可以用一个while循环完成,这个在游戏编程渲染顶点时也经常使用。方法如下:define M 10
define N 2
int i = 0
int a[M][N]
while(i <M*N)
{
a[i/N][i%N] = 0
}
希望能帮到你。
#include <stdio.h>#define N 20
int main( void )
{
int row, col, i, j, k, i1, j1
int num
int array[N][N]
printf( "Please input row and col:\n")
scanf( "%d %d", &row, &col )
num = row * col
for( i = 0i <rowi++ )
for( j = 0j <colj++ )
scanf( "%d", &array[i][j] )
for( i = 0i++ ){
for( j = ij <col - ij++ ){
printf( "%d\t", array[i][j] )
num--
if( num == 0 )
return 0
}
for( k = i + 1k <row - ik++ ){
printf( "%d\t", array[k][j-1])
num--
if( num == 0 )
return 0
}
for( i1 = j - 2i1 >i - 1i1-- ){
printf( "%d\t", array[k-1][i1] )
num--
if( num == 0 )
return 0
}
for( j1 = k - 2j1 >ij1-- ){
printf( "%d\t", array[j1][i] )
num--
if( num == 0 )
return 0
}
}
return 0
}