Hdu1197: Specialized four-digit numbers

Source: Internet
Author: User
Problem descriptionfind 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 1893 (12), and these digits also sum up to 21. but in hexadecimal 2991 is baf16, and 11 + 10 + 15 = 36, so 2991 shoshould
Be rejected by your program.

The next number (2992), however, has digits that sum to 22 in all three representations (including bb016), so 2992 shocould 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 .)

Inputthere is no input for this problem.

Outputyour 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

29922993299429952996299729982999
 

 

// Question

 

#include <iostream>using namespace std;int main(){    int i,sum1,sum2,sum3,n,r;    for(i = 1000; i<9999; i++)    {        n = i;        sum1 = sum2 = sum3 = 0;        while(n)        {            r=n%10;            sum1+=r;            n/=10;        }        n = i;        while(n)        {            r=n%12;            sum2+=r;            n/=12;        }        if(sum1 == sum2)        {            n = i;            while(n)            {                r=n%16;                sum3+=r;                n/=16;            }            if(sum3 == sum1)            cout << i << endl;        }    }    return 0;}

 

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