POJ-1316-Self Numbers
Self Numbers
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 24891 | Accepted: 13910 |
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.
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。
