9*9口诀程序，哪不对啊
 #include "stdio.h"
main()
 {
   int i=1;int j=1;
   int I[9];
   int J[9];
     printf("the projets is 9*9:");
     for (i=1;i<10;i++)
      printf("%d*%d=",I[i],J[j]);
     for (j=1;j<10||i<0;j++,i--)
      printf("%d*%d=",I[i],J[j]);
      getch();
      }

好象好多错误啊！~~

我不用数组
#include<stdio.h>
void main()
{
 int i ,  j;
for(i=1;i<=9;i++)
{
 for(j=1;j<=i,j++)
   printf("%d*%d=%2d",i,j,i*j);
printf("\n");
getch();
}
这也是我书上看来的

你又说说，你这程序哪儿对？

#include<stdio.h>
void main( )
{
 int i,j;
 for(i=1;i<=9;i++)
 {
  for(j=1;j<=i;j++)
   if(i*j<10)
 printf("%d*%d=%d  ",i,j,i*j);
   else
 printf("%d*%d=%d ",i,j,i*j);
  printf("\n");
 }
}

to 3楼：for(j=1;j<=i,j++)//是分号不是逗号，还有程序还差一个}.还有%2d对齐方式不好．
//修改了一下．
#include <stdio.h>
main()
{
   int i,j;
   printf("the projets is 9*9:\n\n");
   for (i=1;i<10;i++)
    {
       for(j=1;j<=i;j++)
       printf("%d*%d=%-4d",i,j,i*j);
       printf("\n");
     }
    getch();
}

我用汇编编了一个，要不要看看？？

好呀,太感谢啦

DATA SEGMENT VA DB 1,2,3,4,5,6,7,8,9 DATA ENDS

STACK SEGMENT PARA STACK DW 20 DUP(0) STACK ENDS

CODE SEGMENT ASSUME CS:CODE,DS:DATA,SS:STACK START: MOV AX,DATA MOV DS,AX MOV SI,0 MOV CX,9H LOP1: PUSH CX MOV CL,VA[SI] MOV CH,0H MOV BL,VA[SI] MOV BH,1H LOP2: PUSH CX MOV AL,BH MUL BL AAM PUSH AX AND BH,0FH ADD BH,30H MOV DL,BH MOV AH,02H INT 21H MOV AH,02H MOV DL,'*' INT 21H AND BL,0FH ADD BL,30H MOV AH,02H MOV DL,BL INT 21H MOV AH,02H MOV DL,'=' INT 21H POP AX PUSH AX AND AH,0FH ADD AH,30H MOV DL,AH MOV AH,02H INT 21H POP AX AND AL,0FH ADD AL,30H MOV DL,AL MOV AH,02H INT 21H MOV AH,02H MOV DL,20H INT 21H AND BL,0FH AND BH,0FH INC BH POP CX LOOP LOP2 MOV AH,02H MOV DL,0DH INT 21H MOV AH,02H MOV DL,0DH INT 21H MOV AH,02H MOV DL,0AH INT 21H POP CX INC SI LOOP LOP1 MOV AH,4CH INT 21H CODE ENDS END START 前几天有事，没来

main()
{int i,j,a[10][10];
 for(i=1;i<10;i++)
  {for(j=1;j<10;j++)
   {a[i][j]=i*j;
    printf("%d*%d=",i,j);
    printf("%d ",a[i][j]);
   }
  }
}