poj1316
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 Input
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
Self Numbers
|
Time Limit: 1000MS |
|
Memory Limit: 10000K |
|
Total Submissions: 7878 |
|
Accepted: 4271 |
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 Input
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
其实水题一道，根本不用理睬一层一层递推，只需要从1到10000全部生成一遍打表就行
1 #include<iostream> 2 using namespace std; 3 int num[100000]={0}; 4 int main() 5 { 6 int i; 7 for(i=1;i<10;i++) 8 { 9 int n=i+i; 10 num[n]=1; 11 } 12 for(i=10;i<100;i++) 13 { 14 int n=i+i/10+i%10; 15 num[n]=1; 16 } 17 for(i=100;i<1000;i++) 18 { 19 int n=i+i/100+i/10%10+i%10; 20 num[n]=1; 21 } 22 for(i=1000;i<10000;i++) 23 { 24 int n=i+i/1000+i/100%10+i/10%10+i%10; 25 num[n]=1; 26 } 27 for(i=1;i<10000;i++) 28 { 29 if(num[i]==0) 30 { 31 cout<<i<<endl; 32 } 33 } 34 return 0; 35 }
