HIT OJ 1087哈工大OJ
Self Numbers
Source : ACM ICPC Mid-Central USA 1998
Time limit : 5 sec
Memory limit : 32 M
Submitted : 1119, Accepted : 448
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
|
|
|
選別法枝を切ってOK
水の問題は#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
bool flag[1000005];
int cal(int n)
{
int ret=n;
while(n)
{
ret=ret+(n%10);
n=n/10;
}
return ret;
}
void init()
{
int i,next;
for(i=1;i<=1000000;i++)
{
if(!flag[i])
{
printf("%d
",i);
next=cal(i);
while(next<=1000000)
{
if(flag[next])
break;
flag[next]=true;
next=cal(next);
}
}
}
}
int main()
{
int i;
init();
return 0;
}
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<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
bool flag[1000005];
int cal(int n)
{
int ret=n;
while(n)
{
ret=ret+(n%10);
n=n/10;
}
return ret;
}
void init()
{
int i,next;
for(i=1;i<=1000000;i++)
{
if(!flag[i])
{
printf("%d
",i);
next=cal(i);
while(next<=1000000)
{
if(flag[next])
break;
flag[next]=true;
next=cal(next);
}
}
}
}
int main()
{
int i;
init();
return 0;
}