Self Numbers
Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6227 Accepted Submission(s): 2728
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, ...
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
|
|
|
9949
9960
9971
9982
9993
|
|
|
Source
尼玛，太简单了，之间就水过去了.....
代码：
#include<cstdio>
#include<cstring>
#define maxn 1000001
/*求个位数之和*/
int work(int n)
{
int sum=;
while(n>){
sum+=n%;
n/=;
}
return sum;
}
bool ans[maxn];
int main(){
int pos;
//freopen("test.out","w",stdout);
memset(ans,,sizeof(ans));
for(int i=;i<maxn;i++){
pos=i+work(i);
if(pos<=&&!ans[pos]) ans[pos]=;
}
for(int i=;i<maxn;i++){
if(!ans[i])printf("%d\n",i);
}
return ;
}