HOJ1087

Self Numbers

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	Source : ACM ICPC Mid-Central USA 1998
	Time limit : 5 sec		Memory limit : 32 M

Submitted : 1443, Accepted : 618

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.

 

题目大意为打印小于1000000以内的自私数

自私数是指可由一个数的各位数字与本身的和组成：例如 2 = 1 + 1；  11 = 1 + 0 + 10； 22 = 2 + 0 + 20；这些都是自私数

 

一开始仿照埃氏筛法的思想，处理一个数字顺带把由这个数字生成的其他所有数字都处理掉。结果总是TLE，附上代码：

 #include<iostream>
  using namespace std;
  
  long MaxSize = 1000000;
  
  long Dc(long Num){
      long D = 0;
      D = Num + (Num%10) + (Num/10)%10 + (Num/100)%10 + (Num/1000)%10 + (Num/10000)%10 +(Num/100000)%10;
      return D;
 }
 
 int main(){
     bool List[MaxSize];
     for(long i = 1;i <= MaxSize;i++){
         if(!List[i])    printf("%d\n",i);
         long no_self = i;
         while(no_self <= MaxSize && !List[no_self]){
             no_self = Dc(no_self);
             if(no_self <= MaxSize)
                 List[no_self] = 1;
         }
     }
     return 0;
 }

 

问题出在17~21行，这段while循环体实质上并没有让外循环for的指标进行非线性变动，即while循环完全是做无用功。这与埃氏筛有本质不同。但观察到，这段代码只需要给出下一个非Self number的序号即可，因此while循环就可以全部摘去，采用在线处理算法的思想，整个算法的复杂度直接将为O(N)，下面为AC代码：

/*This Code is Submitted by mathmiaomiao for Problem 1087 at 2015-08-21 23:21:22*/
#include <iostream>

using namespace std;

bool List[1000000];
int main() {
  long no_self = 0;
  for(long i = 1; i < 1000000; ++i) {
    if(!(List[i])) printf("%ld\n",i);
    no_self = i + (i%10) + (i/10)%10 + (i/100)%10 + (i/1000)%10 + (i/10000)%10 +(i/100000)%10;
    List[no_self] = 1;
  }
  printf("1000000\n");
  return 0;
}