Title Description
Will 1, 2,... Nine digits in total are divided into three groups, each of which consists of three triple digits, and the ratio of the three triple digits is A:B:C. Try to find all three triple digits satisfying the conditions. If there is no solution, output "No!!!".
Input format
Three numbers, A B C.
Output format
Several rows, three digits per row. Arrange in ascending order according to the first number in each line.
Input and Output Samples
Input #1
1 2 3
Output #1
192 384 576 219 438 657 273 546 819 327 654 981
Note/hint
Guarantee A<B<C
Topic Essentials and Analysis
The requirement of the title is that three of the nine numbers are arranged in a group to form three triple digits, and the proportion of A:B:C should be satisfied.
Because these three numbers cannot be selected repeatedly, the minimum three-digit number can be obtained as 123. So we can solve the problem of violence from 123. There are two small details as follows:
- The three digits obtained cannot contain the number 0.
- The three-digit element itself cannot be repeated.
Code
#include<iostream>
#include<algorithm>
using namespace std;
int main()
{
int q,b,c,sum=0,temp;
cin >> q >> b >> c;
for(int i=123;i<999;i++)
{
int a[10]={0},s=0;
if(i/q*c>999)
break;
if(i%q!=0)
continue;
temp = i;//The first number judgment begins.
for(int n=0;n<3;n++)
{
if(temp%10==0||a[temp%10]==1)
{
s=1;
break;
}
a[temp%10]=1;
temp = temp / 10;
}
if(s==1) continue;
int j=i/q*b;
temp = j;//The second number judgement begins.
for(int n=0;n<3;n++)
{
if(a[temp%10]==0&&temp%10!=0)
a[temp%10]=1;
else
{
s=1;
break;
}
temp = temp/10;
}
if(s==1) continue;
int k=i/q*c;
temp = k;//The third number judgment begins.
for(int n=0;n<3;n++)
{
if(a[temp%10]==0 && temp%10!=0)
a[temp%10]=1;
else
{
s=1;
break;
}
temp = temp/10;
}
if(s==1) continue;
else
{
cout << i << " " << j << " " << k <<endl;
sum++;
}
}
if(sum==0)
{
cout << "No!!!" <<endl;
}
return 0;
}