浙大在线评测 1180 Self Numbers

王朝other·作者佚名  2006-01-09
窄屏简体版  字體: |||超大  

Problem:

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

|

|

|

Solution:

#include <iostream>

using namespace std;

// 声明:本代码仅供学习之用,请不要作为个人的成绩提交。

// http://blog.csdn.net/mskia

// email: bitrain@hotmail.com

const long limit = 1000000;

int main( void ) {

bool boo[ limit ];

for ( long i = 0; i < limit; ++i ) {

boo[ i ] = false;

}

for ( long i = 1; i <= limit; ++i ) {

long t = i , r = i;

while ( t > 0 ) {

r += t % 10;

t /= 10;

}

if ( r <= limit ) {

boo[ r - 1 ] = true;

}

}

for ( long i = 0; i < limit; ++i ) {

if ( !boo[ i ] ) {

cout << i + 1 << endl;

}

}

system( "pause" );

return 0;

}

 
 
 
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