我错了，至少有一位是5。。。

根据题目描述，直接构建的模型，肯定可以解决问题，但是程序的效率不一定的最优的，并且那样的解法只要学过编程的人应该都能写出来。如果想要提高效率，就需要通过数学知识改造模型，这固然需要投入精力。如果不愿意投入这样的精力，以后很可能会被淘汰。

试问，现在谁还愿意使用Windows3.x系统的电脑，或者DOS？恐怕没人了吧，因为我们有用起来非常方便的Windows xp。

编程，也就是这个样子。

// 被3整除有5.cpp : Defines the entry point for the console application.
//
#include "stdafx.h"

#include<stdio.h>
void main()
{
    int i,j,k;
    for(i=0;i<=9;i++)
    {
        if((i+5)%3==0)
        {
            printf("%3d %3d ",i*10+5,50+i);
            k+=2;
            if(k%4==0)
                printf("\n");
        }
        for(j=0;j<=9;j++)
        {
            if((i+j+5)%3==0)
            {
                if(i!=j)
                {
                    if((i!=5)&&(j!=5))
                    {
                        printf("%3d %3d ",500+10*i+j,500+10*j+i);
                        k+=2;
                        if(k%4==0)
                            printf("\n");
                       
                        {
                            printf("%3d %3d ",100*i+10*j+5,100*i+50+j);
                            k+=2;
                        }
                        if(k%4==0)
                            printf("\n");
                        
                        {
                            printf("%3d %3d ",100*j+10*i+5,100*j+50+i);
                            k+=2;
                        }
                        if(k%4==0)
                            printf("\n");
                    }
                    if(i==5)
                    {
                        if(k%4==1||k%4==0)
                        {
                            printf("%3d %3d %3d ",550+j,505+10*j,100*j+55);
                            k+=3;
                            if(k%4==0)
                                printf("\n");
                        }
                        else
                        {
                            if(k%4==2)
                            {
                                printf("%3d %3d \n  %3d ",550+j,505+10*j,100*j+55);
                                k+=3;
                            }
                            else
                            {
                                if(k%4==3)
                                {
                                    printf("%3d \n  %3d %3d ",550+j,505+10*j,100*j+55);
                                    k+=3;
                                }
                                
                            }
                        }
                        if(j==5)
                        {}
                    }
                    if(i==j&&i!=5)
                    {
                        if(k%4==1||k%4==0)
                        {
                            printf("%3d %3d %3d ",500+10*i+i,100*i+50+i,100*i+10*i+5);
                            k+=3;
                            if(k%4==0)
                                printf("\n");
                           
                        }
                        else
                        {
                            if(k%4==2)
                            {
                                printf("%3d %3d \n%3d ",500+10*i+i,100*i+50+i,100*i+10*i+5);
                                k+=3;
                            }
                            else
                            {
                                printf("%3d \n%3d %3d ",500+10*i+i,100*i+50+i,100*i+10*i+5);
                                k+=3;
                            }
                        }
                        
                    }
                    if(i==j&&i==5)
                        printf("%3d ",555);
                    
                }            
               
            }
        }
    }
    printf("\n");
}
//可以按你的要求输出~四个数字一行~其中在某些地方加空格就可以了~
//但是其中有数字是重复的~算法方面还是有些问题~

算法是要高效， 可是解决问题也要高效的呀，

御姐的代码写出来 最多用时1分钟，楼主花了几天写出来的所谓的高效算法还是有问题的。
当然了，做题的时候无所谓，花个1年时间解决这题也没关系。

无语，这么简单的程序写这么长~~牛~~

这个东西，枚举法。

3楼就写得很不错啊。你用用她的好了

那能帮我找错嘛？