其实我那个主要是因为要4个一行，不然也不会这么长，而且，程序是有错误的！！！但我查不出来了。。

直接用  n%3==0 不行，这个不是整除了么？

#include <stdio.h>

int main()
{
    int n,i,j,k,c=0;
    for (i=0; i<=9; i++)
        for(j=0;j<=9;j++)
           for(k=0;k<=9;k++)
            {
                  if(((i+j+k)%3==0) && (i==3||j==3||k==3))
               {
                  n=i*100+j*10+k;
                  printf("%d ",n);
                  if(++c>=4)
                  puts(""), c=0;
               }
            }

    return 0;
}
那看看这个吧

把这个改写if(((i+j+k)%3==0) && (i==3||j==3||k==3))
成if((k==3||j==3||i==3)&&((i+j+k)%3==0)))能提高效率

算法的目的，首先必须能解决问题，当然也要考虑效率。

你都不懂就乱评价！

#define PIG_NUMBER_ONE(a,b,c) printf("%d%d%d\n",a,b,c);
#define PIG_NUMBER_TWO(a,b,c) PIG_NUMBER_ONE(a,b,c)\
    PIG_NUMBER_ONE(a,c,b)\
    PIG_NUMBER_ONE(b,a,c)\
    PIG_NUMBER_ONE(b,c,a)\
    PIG_NUMBER_ONE(c,b,a)\
    PIG_NUMBER_ONE(c,a,b)
#define PIG_NUMBER_THREE(a,b) PIG_NUMBER_ONE(a,a,b)\
    PIG_NUMBER_ONE(a,b,a)\
    PIG_NUMBER_ONE(b,a,a)
int main(){   
    for(unsigned int a=0;a<10;a++){
        if(a==3)
            continue;
        for(unsigned int b=a;b<10;b++){
            if(b==3&&b==0)
                continue;
            if((a+b)%3==0){
                PIG_NUMBER_TWO(3,b,a)
            }
        }
    }
    PIG_NUMBER_THREE(0,3)
    PIG_NUMBER_THREE(3,6)
    PIG_NUMBER_THREE(3,9)
    PIG_NUMBER_ONE(3,3,3)
    return 0;
}

[ 本帖最后由 hahayezhe 于 2010-9-24 10:20 编辑 ]

#define PIG_NUMBER_ONE(a,b,c) printf("%s\n%s\n%s\n%s\n%s\n%s\n",PIG_NUMBER_TWO(a,b,c),PIG_NUMBER_TWO(a,c,b)\
    ,PIG_NUMBER_TWO(b,c,a),PIG_NUMBER_TWO(c,b,a),PIG_NUMBER_TWO(b,a,c),PIG_NUMBER_TWO(c,a,b));

#define PIG_NUMBER_TWO(a,b,c) #a#b#c

#define PIG_NUMBER_THREE(a,b,c) printf("%d\n%d\n%d\n",PIG_NUMBER_FOUR(a,b,c),PIG_NUMBER_FOUR(a,c,b),PIG_NUMBER_FOUR(c,a,b));
#define PIG_NUMBER_FOUR(a,b,c) a##b##c
#define PIG_NUMBER_FIVE(a,b,c) printf("%s\n",#a#b#c);
int _tmain(int argc, _TCHAR* argv[])
{
    PIG_NUMBER_ONE(0,6,3)
    PIG_NUMBER_ONE(0,9,3)
    PIG_NUMBER_ONE(1,2,3)
    PIG_NUMBER_ONE(1,5,3)
    PIG_NUMBER_ONE(1,8,3)
    PIG_NUMBER_ONE(2,4,3)
    PIG_NUMBER_ONE(2,7,3)
    PIG_NUMBER_ONE(4,5,3)
    PIG_NUMBER_ONE(4,8,3)
    PIG_NUMBER_ONE(5,7,3)
    PIG_NUMBER_ONE(6,6,3)
    PIG_NUMBER_ONE(6,9,3)
    PIG_NUMBER_THREE(3,3,6)
    PIG_NUMBER_THREE(3,3,9)
    PIG_NUMBER_FIVE(3,3,3)
    PIG_NUMBER_FIVE(0,3,3)
    PIG_NUMBER_FIVE(3,0,3)
    PIG_NUMBER_FIVE(3,3,0)
    return 0;
}

.....

[ 本帖最后由 BlueGuy 于 2010-9-24 10:33 编辑 ]

main()
{
   int count; /*计数器*/
   int i,j,k; /*第几位*/
   int bits[3];/*bits[x]表示第x位上的数字*/
   int m;
   clrscr();
   for(i=0;i<3;i++)
   {
        bits[i]=3;
        for(j=0;j<3;j++)
        {
            if(j==i) continue;
            for(k=0;k<3;k++)
            {
                if(k==i||k==j) continue;
                bits[i]=3;
        for(m=0;m<10;m=m+3)
                {
                     for(bits[j]=0;bits[j]<=(m/2.0);bits[j]++)
             {

             bits[k]=m-bits[j];
             if(bits[j]==3||bits[k]==3) continue;
                         printf("%d",bits[i]*E(i)+bits[j]*E(j)+bits[k]*E(k));
                         count++;
                         if(0==count%4) printf("\n");
                         else printf("\t");
                     }
                }
            }
        }
   }
}
int E(int x)
{
   if(x==0)  return 1;
   else return E(x-1)*10;
}

这个算法有些问题，出现了重复的数字

看不懂楼主的逻辑