POJ 1316 Self Numbers

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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.

Input

No input for this problem.

Output

Write a program to output all positive self-numbers less than 10000 in increasing order, one per line.

Sample Input

Sample Output

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

思路

没啥好说的，用数组进行标记，只要出现过的就为1.

#include<iostream>
#include<cstring>
using namespace std;
int main()
{
    int a[11000];
    int k,sum;
    memset(a,0,sizeof(a));
    for(int i=1;i<10000;i++)
    {
        sum=i;
        k=i;
        while(k)
        {
            sum+=k%10;
            k/=10;
        }
        a[sum]=1;
    }
    for(int i=1;i<10000;i++)
    {
        if(a[i]!=1)
            cout<<i<<endl;
    }
}

Memory: 296K   Time: 16MS

调用函数

#include <iostream>
#include<cstring>
using namespace std;
int self(int n)
{
    int sum=n;
    while(n)
    {
        sum+=n%10;
        n/=10;
    }
    return sum;
}
int main()
{
    int a[15000];
    memset(a,0,sizeof(a));
    for(int i=1;i<10000;i++)
    {
        a[self(i)]=1;
    }
    for(int i=1;i<10000;i++)
    {
        if(a[i]!=1) cout<<i<<endl;
    }
}

Memory: 312K   Time: 0MS