首页 > 代码库 > G - Self Numbers(2.2.1)

G - Self Numbers(2.2.1)

2024-07-12 00:44:45   221人阅读

G - Self Numbers(2.2.1)
Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%I64d & %I64u
Submit Status

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 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
#include<iostream>
#include<cstring>
using namespace std;
int shzi(int k)
{
	int s;
	s=k;
	while(s!=0)
	{
		k=k+s%10;
		s=s/10;
	}


return k;}

int a[10001],n=10000;
int main()
{
	memset(a,0,sizeof(a));
	int i,k;
	for(i=1;i<=10000;i++)
	{
		k=shzi(i);
		if(k<=10000)
		a[k]=1;

	}
	for(i=1;i<=10000;i++)
		if(!a[i])
			cout<<i<<endl;

return 0;
}



联系
我们