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 0Sample Output
4937775
题意：输入n,输入大于n的Smith数.注意素数不是Smith数.
思路：暴力，从n+1开始找，如果是素数就continue，否则算出各个位之和赋给s1，然后分解质因子并求和赋给s2，直到s1=s2.
本鶸：TLE 6次，CE 1次，WA 1次，MLE 1次
AC代码：
#include<iostream> #include<cstdio> #include<cstring> #include<string> #include<cmath> #include<algorithm> using namespace std; bool isprime(int n) { if(n==1) return false; for(int i=2;i*i<=n;i++) if(n%i==0) return false; return true; } int cut(int n) { int rec=0; if(n!=1) while(n) { rec+=n%10; n/=10; } return rec; } int simth(int n) { int rec=0; int i=2; while(i*i<=n) { while(n%i==0) { n/=i; rec+=cut(i); } i++; } rec+=cut(n); return rec; } int main() { int n; while(scanf("%d",&n)!=EOF&&n) { while(1) { n++; if(isprime(n)) continue; int s1=cut(n),s2=simth(n); if(s1==s2) { printf("%d\n",n); break; } } } return 0; }
