Self Numbers
Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 5495 Accepted Submission(s): 2431
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.
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
9
20
31
42
53
64
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<-- a lot more numbers
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9903
9914
9925
9927
9938
9949
9960
9971
9982
9993
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Source
Recommend
Eddy
思路：简单处理一下下即可
代码：
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
using namespace std;
int map[1000010];
int n,flag;
void its_number(int number)
{
int sum = number;
while(number)
{
sum += number % 10;
number /= 10;
}
map[sum] = 1;
}
int main()
{
memset(map,0,sizeof(map));
for(int i = 1;i <= 1000000;i ++)
{
its_number(i);
}
for(int i = 1;i <= 1000000;i ++)
if(map[i] == 0)
printf("%d ",i);
return 0;
}
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
using namespace std;
int map[1000010];
int n,flag;
void its_number(int number)
{
int sum = number;
while(number)
{
sum += number % 10;
number /= 10;
}
map[sum] = 1;
}
int main()
{
memset(map,0,sizeof(map));
for(int i = 1;i <= 1000000;i ++)
{
its_number(i);
}
for(int i = 1;i <= 1000000;i ++)
if(map[i] == 0)
printf("%d ",i);
return 0;
}