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 #include <iostream>using namespace std;bool s[10001]={0};int d(int k){int n=k;while(n) { k+=n%10; n/=10; }return k;}int main(){int i,j;for(i=1;i<=10000;++i) { j=d(i);if(j>10000)continue; s[j]=1; }for(i=1;i<=10000;++i)if(!s[i]) cout<<i<<endl; return 0;}转载于:https://www.cnblogs.com/w0w0/archive/2011/11/21/2256850.html
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