poj2196

Specialized Four-Digit Numbers

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7238		Accepted: 5285

Description

Find and list all four-digit numbers in decimal notation that have the property that the sum of its four digits equals the sum of its digits when represented in hexadecimal (base 16) notation and also equals the sum of its digits when represented in duodecimal (base 12) notation.
For example, the number 2991 has the sum of (decimal) digits 2+9+9+1
= 21. Since 2991 = 1*1728 + 8*144 + 9*12 + 3, its duodecimal
representation is 1893₁₂, and these digits also sum up to 21. But in hexadecimal 2991 is BAF₁₆, and 11+10+15 = 36, so 2991 should be rejected by your program.

The next number (2992), however, has digits that sum to 22 in all three representations (including BB0₁₆),
so 2992 should be on the listed output. (We don't want decimal numbers
with fewer than four digits -- excluding leading zeroes -- so that 2992
is the first correct answer.)

Input

There is no input for this problem

Output

Your
output is to be 2992 and all larger four-digit numbers that satisfy the
requirements (in strictly increasing order), each on a separate line
with no leading or trailing blanks, ending with a new-line character.
There are to be no blank lines in the output. The first few lines of
the output are shown below.

Sample Input

There is no input for this problem

Sample Output

2992
2993
2994
2995
2996
2997
2998
2999
...

Source

Pacific Northwest 2004

 #include <iostream>

 using namespace std;

 int digits(int x)
 {
     int a = x/;
     int sum = ;
     sum +=a;
     a = x%/;
     sum +=a;
     a = x%%/;
     sum +=a;
     a = x%%%;
     sum +=a;
     return sum;
 }
 int digits_12(int x)
 {
     int num_12[];
     for(int i = ;i<;i++)
     {
         num_12[i]=x%;
         x=x/;
     }
     return num_12[]+num_12[]+num_12[]+num_12[];
 }
 int digits_16(int x)
 {
     int num_16[];
     for(int i = ;i<;i++)
     {
         num_16[i]=x%;
         x=x/;
     }
     return num_16[]+num_16[]+num_16[]+num_16[];
 }
 int main()
 {

     int num_16[];
     int x = ;
     for(int i=x;i<=;i++)
     {
         int sum = digits(i);
         int sum_12 = digits_12(i);
         int sum_16 = digits_16(i);
         if(sum ==sum_12&&sum==sum_16)
         {
             cout<<i<<endl;
         }
     }
     return ;
 }