目的 打印如下 "9 宮格"
1*1 = 1 1*2 = 2 1*3 = 3
2*1 = 2 2*2 = 4 2*3 = 6
3*1 = 3 3*2 = 6 3*3 = 9
4*1 = 4 4*2 = 8 4*3 = 12
5*1 = 5 5*2 = 10 5*3 = 15
6*1 = 6 6*2 = 12 6*3 = 18
7*1 = 7 7*2 = 14 7*3 = 21
8*1 = 8 8*2 = 16 8*3 = 24
9*1 = 9 9*2 = 18 9*3 = 27
1*4 = 4 1*5 = 5 1*6 = 6
2*4 = 8 2*5 = 10 2*6 = 12
3*4 = 12 3*5 = 15 3*6 = 18
4*4 = 16 4*5 = 20 4*6 = 24
5*4 = 20 5*5 = 25 5*6 = 30
6*4 = 24 6*5 = 30 6*6 = 36
7*4 = 28 7*5 = 35 7*6 = 42
8*4 = 32 8*5 = 40 8*6 = 48
9*4 = 36 9*5 = 45 9*6 = 54
1*7 = 7 1*8 = 8 1*9 = 9
2*7 = 14 2*8 = 16 2*9 = 18
3*7 = 21 3*8 = 24 3*9 = 27
4*7 = 28 4*8 = 32 4*9 = 36
5*7 = 35 5*8 = 40 5*9 = 45
6*7 = 42 6*8 = 48 6*9 = 54
7*7 = 49 7*8 = 56 7*9 = 63
8*7 = 56 8*8 = 64 8*9 = 72
9*7 = 63 9*8 = 72 9*9 = 81
#include <stdio.h>
int main(void)
{
int i, j;
for (i=1; i<=9; i++)
{
for (j=1; j<=3; j++)
{
printf("%d*%d = %2d\t",i ,j ,i*j);
}
printf("\n");
}
printf("\n");
for (i=1; i<=9; i++)
{
for (j=4; j<=6; j++)
{
printf("%d*%d = %2d\t",i ,j ,i*j);
}
printf("\n");
}
printf("\n");
for (i=1; i<=9; i++)
{
for (j=7; j<=9; j++)
{
printf("%d*%d = %2d\t",i ,j ,i*j);
}
printf("\n");
}
getchar();
return 0;
}
請問可否化簡for的數量?
該如何精簡程式?
thx
[此贴子已经被作者于2007-8-28 6:32:46编辑过]
