D - Specialized Four-Digit Numbers

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 1893 ₁₂, and these digits also sum up to 21. But in hexadecimal 2991 is BAF ₁₆, 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 BB0 ₁₆), 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
...


依旧是submit failed，以下为我写的代码

#include <iostream>
using namespace std;
int f(int n,int x)
{
    int i=0,j,a[10],m=0;
    while(n!=0)    {    
        a[i++]=n%x;    
        n=n/x;
    }    
    for(j=0;j<i;j++)m+=a[j];
    return m;
}
int main()
{
    int n,i,p,q;
    for(i=2992;i<10000;i++){
        n=i/1000+i/100%10+i/10%10+i%10;
        p=f(i,16);
        q=f(i,12);
        if(p==n&&q==n)cout<<i<<endl;
    }
    system("pause");
    return 0;
}

以下是提交成功AC的代码

#include <iostream>
using namespace std;
int f(int n,int x)
{
    int a,m=0;
    while(n!=0)    {    
        a=n%x;    
        n=n/x;
        m+=a;
    }    
    return m;
}
int main()
{
    int n,i;
    for(i=2992;i<10000;i++){
        n=i/1000+i/100%10+i/10%10+i%10;
        if(f(i,16)==n&&f(i,12)==n)cout<<i<<endl;
    }
    //system("pause");
    return 0;
}

发现我写的代码似乎太啰嗦了！！！麻烦！！许多东西完全可以省略！！！！