poj1142 Smith Numbers
来源:互联网 发布:windows 版 kinect 编辑:程序博客网 时间:2024/06/05 20:35
Poj1142 Smith Numbers
Smith Numbers
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 13854 Accepted: 4716
Description
While skimming his phone directory in 1982, Albert Wilansky, a mathematician of Lehigh University,noticed that the telephone number of his brother-in-law H. Smith had the following peculiar property: The sum of the digits of that number was equal to the sum of the digits of the prime factors of that number. Got it? Smith’s telephone number was 493-7775. This number can be written as the product of its prime factors in the following way:
4937775= 3*5*5*65837
The sum of all digits of the telephone number is 4+9+3+7+7+7+5= 42,and the sum of the digits of its prime factors is equally 3+5+5+6+5+8+3+7=42. Wilansky was so amazed by his discovery that he named this kind of numbers after his brother-in-law: Smith numbers.
As this observation is also true for every prime number, Wilansky decided later that a (simple and unsophisticated) prime number is not worth being a Smith number, so he excluded them from the definition.
Wilansky published an article about Smith numbers in the Two Year College Mathematics Journal and was able to present a whole collection of different Smith numbers: For example, 9985 is a Smith number and so is 6036. However,Wilansky was not able to find a Smith number that was larger than the telephone number of his brother-in-law. It is your task to find Smith numbers that are larger than 4937775!
Input
The input file consists of a sequence of positive integers, one integer per line. Each integer will have at most 8 digits. The input is terminated by a line containing the number 0.
Output
For every number n > 0 in the input, you are to compute the smallest Smith number which is larger than n,and print it on a line by itself. You can assume that such a number exists.
Sample Input
4937774
0
Sample Output
4937775
Source
Mid-Central European Regional Contest 2000
题意 大于n满足你的各个位数之和等于质因子各位数之和。
题解 暴力过。
#include<cstdio>int fun(long long a){ int sum=0; while(a>0) { sum+=a%10; a/=10; } return sum; }bool prime(long long a){ int flag=1; if (a==1) return false; if(a==2) return true; for(int i=2;i*i<a+1;i++) if(a%i==0) { flag=0; break; } if(flag) return true; else return false;}int cnt(long long a){ if(prime(a)) return fun(a); else { for(int i=2;i*i<a+1;i++) { if(a%i==0) return cnt(i)+cnt(a/i); } }}int main(){ long long a; while(scanf("%lld",&a)!=EOF&&a) { while(a++) { int sum=fun(a); if(!prime(a)&&fun(a)==cnt(a)) break; } printf("%lld\n",a); } return 0; }
这时间倒也不是很多,79ms
- hdu1333/poj1142-Smith Numbers
- poj1142 Smith Numbers
- poj1142 Smith Numbers
- poj1142 Smith Numbers
- POJ1142——Smith Numbers
- Smith Numbers(Poj1142)(质因数分解+素数判定)
- POJ1142 HDU1333 ZOJ1133 Smith Numbers【质因数分解+素数判定+数位之和】
- Smith Numbers
- Smith Numbers
- Smith Numbers
- POJ 1142 Smith Numbers
- 1027: Smith Numbers
- zoj 1133 Smith Numbers
- zoj 1133 - Smith Numbers
- uva 10042 smith numbers
- POJ 1142 Smith Numbers
- poj 1142 Smith Numbers
- Smith Numbers hdu1333 素数
- 洛谷1006 传纸条
- liunx 优化及性能调优
- mysql
- 剑指offer面试题目:数组中只出现一次的数字
- Android 的事件分发机制
- poj1142 Smith Numbers
- WebView与H5那些事儿(=)
- 内存屏障
- MaterialDesign(一)
- DB2查询char型字段问题
- [分治] 51nod算法马拉松27 A.合法括号子段
- 使用Nodejs发送邮件
- 例说数据结构&STL(十二)——iterator
- 算法导论思考题6-2:d叉堆