Self Numbers
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
1 #include<iostream> 2 using namespace std; 3 4 long MaxSize = 1000000; 5 6 long Dc(long Num){ 7 long D = 0; 8 D = Num + (Num%10) + (Num/10)%10 + (Num/100)%10 + (Num/1000)%10 + (Num/10000)%10 +(Num/100000)%10; 9 return D; 10 } 11 12 int main(){ 13 bool List[MaxSize]; 14 for(long i = 1;i <= MaxSize;i++){ 15 if(!List[i]) printf("%d ",i); 16 long no_self = i; 17 while(no_self <= MaxSize && !List[no_self]){ 18 no_self = Dc(no_self); 19 if(no_self <= MaxSize) 20 List[no_self] = 1; 21 } 22 } 23 return 0; 24 }
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:
1 /*This Code is Submitted by mathmiaomiao for Problem 1087 at 2015-08-21 23:21:22*/ 2 #include <iostream> 3 4 using namespace std; 5 6 bool List[1000000]; 7 int main() { 8 long no_self = 0; 9 for(long i = 1; i < 1000000; ++i) { 10 if(!(List[i])) printf("%ld ",i); 11 no_self = i + (i%10) + (i/10)%10 + (i/100)%10 + (i/1000)%10 + (i/10000)%10 +(i/100000)%10; 12 List[no_self] = 1; 13 } 14 printf("1000000 "); 15 return 0; 16 }