POJ-1316-Self Numbers

谁借莪１个温暖的怀抱￠ 2021-09-30 13:54 328阅读 0赞

Self Numbers

<table> 
 <tbody> 
 <tr> 
 <td>Time Limit:&nbsp;1000MS</td> 
 <td>&nbsp;</td> 
 <td>Memory Limit:&nbsp;10000K</td> 
 </tr> 
 <tr> 
 <td>Total Submissions:&nbsp;24891</td> 
 <td>&nbsp;</td> 
 <td>Accepted:&nbsp;13910</td> 
 </tr> 
 </tbody> 
</table>

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
 
 问题分析:
 题目要求输出1-10000之间的自我数。
 显然该题的input size为10000，考虑到只给到了1s的运行时间，所以不能直接用暴力法去破解。
 
 经过仔细观察，如果一个数a是另外一个数b的next number，那么b<=a-36(9999的next number为9999+36),所以我们可以减少算法的操作单元
 
 所以写出如下代码：

1 #include<iostream>
 2 using namespace std;
 3 int main()
 4 {
 5 int i;
 6 for(i=1;i<=10000;i++)
 7 {
 8 int j;
 9 for(j=i-36;j<i;j++)
 10 {
 11 if((j+j%10+(j/10)%10+(j/100)%10+(j/1000)%10)==i)
 12 {
 13 break;
 14 }
 15 }
 16 if(j==i)cout<<i<<endl;
 17 }
 18 return 0;
 19 }

** **

即将本来要从1开始的循环缩短为i-36，这样o(n^2)的复杂度就变成了o(n)的复杂度。

我们可以在做进一步的思考，为什么不用一个used数组将1-10000之间所有数的next number记录下来呢，这样就可以一个简单的循环直接输出结果就行了。

这样算法进一步得到改善，得到如下代码：

1 #include<iostream>
 2 using namespace std;
 3 int used[10000+36]={
 0};
 4 int main()
 5 {
 6 int i;
 7 for(i=1;i<=10000;i++)
 8 {
 9 used[i+i%10+(i/10)%10+(i/100)%10+(i/1000)%10+(i/10000)%10]=1;
 10 }
 11 for(i=1;i<10000;i++)
 12 {
 13 if(used[i]==0) cout<<i<<endl; 
 14 }
 15 return 0;
 16 }

# 如下为运行后poj给出的结果 第一行为法二。 #

![1300037-20180804213049532-1471168878.png][]

转载于:https://www.cnblogs.com/bo2000/p/9420158.html

[1300037-20180804213049532-1471168878.png]: /images/20210724/e5e801803c6044c3ac358c8f357cadf5.png