import java.util.HashSet;
import java.util.Set;

public class Main {
public static void main(String[] args){
for(int i=123;i<330;i++){
int a = i;
int b = i * 2;
int c = i * 3;
if(confirm(String.valueOf(a), String.valueOf(b), String.valueOf(c))){
System.out.println(a+" "+b+" "+c);
}
}
}

private static boolean confirm(String a, String b, String c){
Set<String> set = new HashSet<String>();
for(int i=0;i<3;i ++){
set.add(a.substring(0, 1));
set.add(b.substring(0, 1));
set.add(c.substring(0, 1));
a = a.substring(1);
b = b.substring(1);
c = c.substring(1);
}
if(set.size() == 9 && !set.contains("0")){
return true;
}else{
return false;
}
}
}

#include<bits/stdc++.h>
int main(){
    char s[10];
    bool flag[10]={0},flag2=0;
    for(int i=123;i<=329;++i){
        sprintf(s+1,"%d%d%d",i,i*2,i*3);
        flag[0]=1;//确保0不出现
        for(int j=1;j<=9;++j) if(1^flag[s[j]-'0']) flag[s[j]-'0']=1;else flag2=1;//用异或保证每个数字只出现一次
        if(!(flag2)) printf("%d %d %d\n",i,i*2,i*3);
        memset(flag,0,sizeof(flag));flag2=0;
    }
}

java来一波，感觉想复杂了。

public class FillNumber {

    public static void main(String[] args) {
        for (int i = 100; i <= 334; i ++) {
            int num1 = i;
            int num2 = i * 2;
            int num3 = i * 3;
            if(isConform(num1,num2,num3)) {
                System.out.println(num1 + " " + num2 + " " + num3);
            }
        }
        

    }
    public static boolean isConform(int a, int b, int c) {
        int[] arr = new int[9];
        boolean[] check = new boolean[9];
        for (int i = 0; i < 9; i ++) {
            arr[i] = i + 1;
            check[i] = true;
        }
        int a1 = a % 10; int a2 = (a / 10) % 10; int a3 = a / 100;
        int b1 = b % 10; int b2 = (b / 10) % 10; int b3 = b / 100;
        int c1 = c % 10; int c2 = (c / 10) % 10; int c3 = c / 100;
        for (int i = 0; i < arr.length; i ++) {
            if (check[i]) {
                if (arr[i] == a1 || arr[i] == a2 || arr[i] == a3) {
                    check[i] = false;
                    continue;
                }
                if (arr[i] == b1 || arr[i] == b2 || arr[i] == b3) {
                    check[i] = false;
                    continue;
                }
                if (arr[i] == c1 || arr[i] == c2 || arr[i] == c3) {
                    check[i] = false;
                    continue;
                }
            }
        }
        boolean flag = true;
        for (int i = 0; i < check.length; i ++) {
            if (check[i]) {
                flag = false;
            }
        }
        if(flag) {
            return true;
        }
        return false;
    }   
}

//直接输出答案233~~
#include <iostream>
using namespace std;
int main()
{
cout<<"192 384 576"<<endl;
cout<<"219 438 657"<<endl;
cout<<"273 546 819"<<endl;
cout<<"327 654 981"<<endl;
return 0;
}

短短短
cpp
#include<stdio.h>
int main(){
printf("192 384 576\n219 438 657\n273 546 819\n327 654 981");
return 0;
}


//不解释
#include<iostream>
using namespace std;
int main (){
cout<<192<<' '<<384<<' '<<576<<endl;
cout<<219<<' '<<438<<' '<<657<<endl;
cout<<273<<' '<<546<<' '<<819<<endl;
cout<<327<<' '<<654<<' '<<981<<endl;
return 0;
}

稳到没毛病！（不是我做的）
、

#include<iostream>
#include<algorithm>
#include<string>
using namespace std;
int t[10];
int b1,b2,b3,c,m=0;
int main()
{
for(int i=1;i<=9;i++)
for(int j=1;j<=9;j++)
for(int k=1;k<=9;k++)
{
if(i==j||i==k||j==k)continue;
for(int q=1;q<=9;q++)
t[q]=0;
t[i]++;
t[j]++;
t[k]++;
b1=i*100+j*10+k;
b2=b1*2;
if(b2>999)break;
t[b2/100]++;
t[(b2%100)/10]++;
t[b2%10]++;
b3=b1*3;
if(b3>999)break;
t[b3/100]++;
t[(b3%100)/10]++;
t[b3%10]++;
c=0;
for(int x=1;x<=9;x++)
{
if(t[x]==1) c++;
}
if(c==9)cout<<b1<<" "<<b2<<" "<<b3<<endl;
}
return 0;
}

#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
int main()
{
//freopen("digital.in","r",stdin);
//freopen("digital.out","w",stdout)
cout<<"192 384 576"<<endl;
cout<<"219 438 657"<<endl;
cout<<"273 546 819"<<endl;
cout<<"327 654 981"<<endl;
return 0;
}

.......
```cpp
#include<cstdio>
#include<cstring>
char s[10];
bool b[10];
bool pd(int x)
{
sprintf(s,"%d%d%d",x,x*2,x*3);
memset(b,0,sizeof(b));
for(int i=0;s[i];i++)
{
if(s[i]=='0')return false;
if(b[s[i]-'0'])return false;
b[s[i]-'0']=true;
}
return true;
}
int main()
{
for(int i=123;i*3<=999;i++)
if(pd(i))
printf("%d %d %d\n",i,i*2,i*3);
return 0;
}

var i,a,b,c:longint;
    ch,ch1,ch2:string;

function max(x,y,z:longint):longint;
var i,j,t:longint;
  a:array[1..3]of longint;
begin
  t:=0;
  a[1]:=x; a[2]:=y; a[3]:=z;
  for i:=1 to 3-1 do
  for j:=i+1 to 3 do if a[i]<a[j] then begin  t:=a[j]; a[j]:=a[i]; a[i]:=t; end;
  exit(a[1]);
end;
function f(s:string):boolean;
var c:char;
begin
    for c:='1'to'9'do
        if pos(c,s)=0 then exit(false);
    exit(true);
end;
begin
  readln(a,b,c); a:=1; b:=2; c:=3;
    for i:=123 to 1000 div max(a,b,c)   do
    begin
        str(i*a,ch);
        str(i*b,ch1);
        str(i*c,ch2);
        ch:=ch+ch1+ch2;
        if f(ch)then writeln(i*a,' ',i*b,' ',i*c);
    end;
end.

打表万岁！！！
先暴力找出来，然后写打表
```cpp
#include<iostream>
using namespace std;

int main()
{
std::ios::sync_with_stdio(false);
cout<<"192 384 576"<<endl;
cout<<"219 438 657"<<endl;
cout<<"273 546 819"<<endl;
cout<<"327 654 981"<<endl;
}
```

Var
     i,j,k,l,sum,num,num1,sum1:longint;
     a,f:array[0..9]of integer;
     ch:boolean;
Begin
     for i:=1 to 9 do
          for j:=1 to 9 do
               for k:=1 to 9 do
               if (i<>j) and (j<>k) and (i<>k) then
               begin
                    fillchar(a,sizeof(a),0);
                    a[1]:=i; a[2]:=j; a[3]:=k;
                    num:=i*100+j*10+k;
                    if (num*2>999) or (num*3>999) then
                         continue;
                    num1:=num*2; sum:=num1;
                    a[4]:=num1 mod 10; num1:=num1 div 10;
                    a[5]:=num1 mod 10; num1:=num1 div 10;
                    a[6]:=num1 mod 10; num1:=num1 div 10;
                    num1:=num*3; sum1:=num1;
                    a[7]:=num1 mod 10; num1:=num1 div 10;
                    a[8]:=num1 mod 10; num1:=num1 div 10;
                    a[9]:=num1 mod 10; num1:=num1 div 10;
                    ch:=true;
                    fillchar(f,sizeof(f),0);
                    for l:=1 to 9 do
                         inc(f[a[l]]);
                    for l:=1 to 9 do
                         if f[l]<>1 then
                              ch:=false;
                    if ch then
                         writeln(num,' ',sum,' ',sum1);
               end;
     readln;
End.

如果今年的题目像这样简单就好，，水题水过。。。

#include<cstdio>
int main(){printf("192 384 576\n219 438 657\n273 546 819\n327 654 981");}
打表
93Bytes

#include <stdio.h>
#include<iostream>
using namespace std;
void Getnum(int x, int a[], int n){ //计算个十百位上的数并存到数组中去
while (x != 0){
a[n - 1] = x % 10;
x = x / 10;
n--;
}
}
bool Have_same(int a[]){//判断一个数组中有没有相同的数
if (a[0] != a[1] && a[0] != a[2] && a[1] != a[2]){
return true;
}
else return false;
}
bool Is_true(int a[], int b[]) { //比较3个数各位置上的数是否相同并查找是否包括0，0不在这几个数中
for (int i = 0; i < 3; i++){
for (int j = 0; j < 3; j++){
if (a[i] == b[j]||a[i]==0||b[j]==0){ return false; }
}
}
return true;
}
int main() //123456789一共9个数，满足条件构造3个数
{
int num1, num2, num3; //存放3个不同的数
int num11[3], num22[3], num33[3]; //3个数组存放每个数个十百位上的数
for (int i = 123; i <= 329; i++){ //从最小的数123循环到最大的数329就可以找到所有的数
if ((i % 10) != 0){
num1 = i;
num2 = 2 * i;
num3 = 3 * i;
Getnum(num1, num11, 3);
Getnum(num2, num22, 3);
Getnum(num3, num33, 3);
if (Is_true(num11, num22) && Is_true(num11, num33) && Is_true(num33, num22) && Have_same(num11) && Have_same(num22) && Have_same(num33)) {
cout << num1 << ' ' << num2 << ' ' << num3 << endl;
}
}
}
system("pause");
}

#include <stdio.h>
#include<iostream>
using namespace std;
void Getnum(int x, int a[], int n){ //计算个十百位上的数并存到数组中去
while (x != 0){
a[n - 1] = x % 10;
x = x / 10;
n--;
}
}
bool Have_same(int a[]){//判断一个数组中有没有相同的数
if (a[0] != a[1] && a[0] != a[2] && a[1] != a[2]){
return true;
}
else return false;
}
bool Is_true(int a[], int b[]) { //比较3个数各位置上的数是否相同并查找是否包括0，0不在这几个数中
for (int i = 0; i < 3; i++){
for (int j = 0; j < 3; j++){
if (a[i] == b[j]||a[i]==0||b[j]==0){ return false; }
}
}
return true;
}
int main() //123456789一共9个数，满足条件构造3个数
{
int num1, num2, num3; //存放3个不同的数
int num11[3], num22[3], num33[3]; //3个数组存放每个数个十百位上的数
for (int i = 123; i <= 329; i++){ //从最小的数123循环到最大的数329就可以找到所有的数
if ((i % 10) != 0){
num1 = i;
num2 = 2 * i;
num3 = 3 * i;
Getnum(num1, num11, 3);
Getnum(num2, num22, 3);
Getnum(num3, num33, 3);
if (Is_true(num11, num22) && Is_true(num11, num33) && Is_true(num33, num22) && Have_same(num11) && Have_same(num22) && Have_same(num33)) {
cout << num1 << ' ' << num2 << ' ' << num3 << endl;
}
}
}
system("pause");
}

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

bool isvalid(const int &val,bool vis[]){
int num=val;
int a,b,c;
a=num/100;
num-=(a*100);
b=num/10;
c=num%10;
if(a==b||a==c||b==c)return false;
if(a==0||b==0||c==0)return false;
vis[a]=1;
vis[b]=1;
vis[c]=1;
return true;
}

int main(){
int flag=0;
for(int i=100;i<=333;++i){
bool vis[10]={0};
if(isvalid(i,vis)&&isvalid(2*i,vis)&&isvalid(3*i,vis)){
for(int j=1;j<10;++j){
if(!vis[j]){
flag=1;
break;
}
}
if(!flag)cout<<i<<" "<<2*i<<" "<<3*i<<"\n";
flag=0;
}
}
//system("pause");
return 0;
}

// int main(){
// cout<<192<<" "<<384<<" "<<576<<"\n"
// <<219<<" "<<438<<" "<<657<<"\n"
// <<273<<" "<<546<<" "<<819<<"\n"
// <<327<<" "<<654<<" "<<981;
// system("pause");
// return 0;
// }

纯数学问题
不难看出三个数相加一定是6的倍数，可以枚举所有三个数的和可能的情况

#include<iostream>
#include<cstring>
using namespace std;
int x[4];bool p[10];
int main()
{
    int a=606,b=2700,l,i,k;
    for(i=a;i<=b;i+=6)
    {
        memset(p,false,sizeof(p));
        x[1]=i/6;
        x[2]=i/3;
        x[3]=i/2;
        for(k=1;k<=3;k++)
         {
          p[x[k]/100]=true;
          p[(x[k]/10)%10]=true;
          p[x[k]%10]=true;
         }
         for(l=1;l<=9;l++)
          if(!p[l])
           break;
        if(l==10)
        {
            cout<<x[1]<<" "<<x[2]<<" "<<x[3]<<endl;
        }
    }
    return 0;
}

#include<iostream>
#include<string>
#include<set>
using namespace std;

void addto_set(set<int>& s, int num)
{
    int t = 0;
    while (num > 0)
    {
        t = num % 10;
        s.insert(t);
        num = num / 10;
    }
}
int main()
{

    //freopen("data.in", "r", stdin);
    //freopen("data.out", "w", stdout);
    
    
    int fir = 0;
    int sec = 0;
    int thir = 0;
    set<int> s;
    set<int>::iterator iter;
    for (fir = 100; fir <= 999; fir++)
    for (sec = 100; sec <= 999;sec++)
    for (thir = 100; thir <= 999; thir++)
    {
        if (2 * fir == sec && 3 * fir == thir)
        {
            s.clear();
            addto_set(s, fir);
            addto_set(s, sec);
            addto_set(s, thir);
            if (s.size() == 9 && (iter=s.find(0))==s.end())
                cout << fir << " " << sec << " " << thir << " " << endl;
            else
                continue;
            
        }           
        else
            continue;   
    }
    
    
    return 0;
}

终极写法。。我觉得我没法精简下去了

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int comp(const void *a, const void *b){return *(char*)a-*(char*)b;} 
int main(){
    int i=100,j;
    char a[12];
w:while(j=1,i++<333){
        sprintf(a,"%d%d%d",i,i*2,i*3);
        qsort(a, 9, sizeof(*a), comp); 
        if(a[0]=='0')continue;
        while(j<9)
            if(a[j]==a[j++-1])goto w;
        printf("%d %d %d\n",i,i*2,i*3);
    }
}

//简单粗暴，直接有效

#include<iostream>
using namespace std;
bool pan(int a,int b,int c){
    bool num[9]={true,true,true,true,true,true,true,true,true};
    int E[9];
    E[0]=a%10;
    E[1]=a/10%10;
    E[2]=a/100%10;
    E[3]=b%10;
    E[4]=b/10%10;
    E[5]=b/100%10;
    E[6]=c%10;
    E[7]=c/10%10;
    E[8]=c/100%10;
    for(int i=0;i<=8;i++){
        if(E[i]==0)return false;
        if(num[E[i]-1])num[E[i]-1]=false;
        else {
            return false;
        } 
    }
    return true;
}
int main(){
    int a,b,c;
    for(int i=123;i<=987;i++){
        a=i*1;
        b=i*2;
        c=i*3;
        if(b<1000&&c<1000&&pan(a,b,c))cout<<a<<" "<<b<<" "<<c<<endl;
    }
}