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, ...
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.
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
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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
尼玛，太简单了，之间就水过去了.....
代码：
1 #include<cstdio> 2 #include<cstring> 3 #define maxn 1000001 4 /*求个位数之和*/ 5 int work(int n) 6 { 7 int sum=0; 8 while(n>0){ 9 sum+=n%10; 10 n/=10; 11 } 12 return sum; 13 } 14 bool ans[maxn]; 15 int main(){ 16 int pos; 17 //freopen("test.out","w",stdout); 18 memset(ans,0,sizeof(ans)); 19 for(int i=1;i<maxn;i++){ 20 pos=i+work(i); 21 if(pos<=1000000&&!ans[pos]) ans[pos]=1; 22 } 23 for(int i=1;i<maxn;i++){ 24 if(!ans[i])printf("%d ",i); 25 } 26 return 0; 27 }