Specialized Four-Digit Numbers(进制转换)

Specialized Four-Digit Numbers

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4978    Accepted Submission(s): 3597

Problem 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 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<cstdio>
#include<iostream>
#include<cstring>
const int maxn=1000;
using namespace std;
int t[maxn],A[maxn];
char str1[maxn],str2[maxn];
int n,m;
void solve(char *str1,char *str2,int n,int m)
{
    int i,len,k;
    len=strlen(str1);
    for(i=len;i>=0;i--)
    {
        t[len-1-i]=str1[i]-(str1[i]<58?48:str1[i]<97?55:61);
        //小于58说明是48-57是数字
        //小于97是65-90为大写英文字母
        //大于97是97-122为小写英文字母！
        //同时也说明了0-9存的是数字
        //10-25存的是大写字母
        //26-61存的是小写字母
    }
    for(k=0;len;)
    {
        for(i=len;i>=1;i--)
        {
            t[i-1]+=t[i]%m*n;
            t[i]/=m;
        }
        A[k++]=t[0]%m;
        t[0]/=m;
        while(len>0&&!t[len-1])
        {
            len--;
         }
    }
    str2[k]=NULL;
    for(i=0;i<k;i++)
    {
        str2[k-1-i]=A[i]+(A[i]<10?48:A[i]<36?55:61);
    }
}
int main()
{
    int i,j,sum1,sum2,sum3;
    for(i=1000;i<=9999;i++)
    {
        sum1=sum2=sum3=0;
        sprintf(str1,"%d",i);
        for(j=0;j<strlen(str1);j++)
        {
            sum1+=str1[j]-'0';
        }
        solve(str1,str2,10,16);
        for(j=0;j<strlen(str2);j++)
        {
            if(str2[j]>='A'&&str2[j]<='Z')
            {
                sum2+=str2[j]-'A'+10;
            }
            else
            {
                sum2+=str2[j]-'0';
            }
        }
        solve(str1,str2,10,12);
        for(j=0;j<strlen(str2);j++)
        {
            if(str2[j]>='A'&&str2[j]<='Z')
            {
                sum3+=str2[j]-'A'+10;
            }
            else
            {
                sum3+=str2[j]-'0';
            }
        }
        if(sum1==sum2&&sum1==sum3)
        {
            cout<<str1<<endl;
        }
    }
}