Self Numbers

 

Self Numbers

Time Limit : 20000/10000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 34   Accepted Submission(s) : 16

Problem Description

In 1949 the Indian mathematician D.R. Kaprekar discovered a class of numbers called self-numbers. For any positive integer n, define d(n) to be n plus the sum of the digits of n. (The d stands for digitadition, a term coined by Kaprekar.) For example, d(75) = 75 + 7 + 5 = 87. Given any positive integer n as a starting point, you can construct the infinite increasing sequence of integers n, d(n), d(d(n)), d(d(d(n))), .... For example, if you start with 33, the next number is 33 + 3 + 3 = 39, the next is 39 + 3 + 9 = 51, the next is 51 + 5 + 1 = 57, and so you generate the sequence 33, 39, 51, 57, 69, 84, 96, 111, 114, 120, 123, 129, 141, ... The number n is called a generator of d(n). In the sequence above, 33 is a generator of 39, 39 is a generator of 51, 51 is a generator of 57, and so on. Some numbers have more than one generator: for example, 101 has two generators, 91 and 100. A number with no generators is a self-number. There are thirteen self-numbers less than 100: 1, 3, 5, 7, 9, 20, 31, 42, 53, 64, 75, 86, and 97. Write a program to output all positive self-numbers less than or equal 1000000 in increasing order, one per line.

 

 

Sample Output

1 3 5 7 9 20 31 42 53 64 | | <-- a lot more numbers | 9903 9914 9925 9927 9938 9949 9960 9971 9982 9993 | | |

 

 

Source

Mid-Central USA 1998

 

#include <stdio.h>
#include <string.h>
int sum[1000005]={0};
int All_sum(int n)
{
    if (n<10)
        return n;
    else
        return (n%10)+All_sum(n/10);
 }

void num(int i,int n)
{
    int j,k=i-9*n,tmp;
    if(k<0)
        k=1;
    while(1)
    {
        tmp=k;
        if(k>i)
            return 0;
        tmp+=All_sum(k);
        if(tmp==i)
        {sum[tmp]+=1;return 0;}
        k++;
    }
    return 0;
}

int main()
{
    int i,n,Len,k,a,j;
    for(i=1;i<=1000000;i++)
    {
        if(i<10)n=1;else if(i<100)n=2; else if(i<1000)n=3; else if(i<10000)n=4; else if(i<100000)n=5; else if(i<1000000)n=6;
        if(sum[i]==0)
           {
               num(i,n);
           }
        if(sum[i]==0)
           printf("%d\n",i);
    }
    return 0;
}