1021: Self Numbers
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Status In/Out TIME Limit MEMORY Limit Submit Times Solved Users JUDGE TYPE
stdin/stdout 10s 16384K 1810 687 Standard
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
|
|
|
我的代码：
#include <iostream>
using namespace std;
int isnotself[1000000]={0};
int Bitcout(int number,int bitnumber) //用来计算一个数有多少位,bitnumber用来统计一个数有多少位
{
if (number/10<1)
return bitnumber;
else
number=number/10;
return Bitcout(number,bitnumber)+1;
}
void d(int n)
{
if (n<=1000000)
{
int btn=Bitcout(n,1);
int bitshu[8]={0};
for (int i=1,n1=n;i<=btn&&n1!=0;i++)
{
bitshu[i]=n1%10;
n1=n1/10;
}
for (int i=1;i<=btn;i++)
n+=bitshu[i];
isnotself[n]=1;
d(n);
}
else
return;
}
int main()
{
for (int i=1;i<=1000000;i++)
if(isnotself[i]==0)
d(i);
for (int i=1;i<=1000000;i++)
if(isnotself[i]==0)
cout<<i<<endl;
return 0;
}
当把1000000换成100000时答案没什么问题，用1000000就是超时，请强人简化一下时间复杂度，谢谢
