Smith Numbers
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 15619 Accepted: 5226
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
寻找大于输入数并且最接近的Smith Numbers。
枚举大于给定n的所有合数（题目中说： a prime number is not worth being a Smith number）是否为Smith Number。
对于枚举的每一个合数，我们首先计算当前数的各个位上的数字和，接下来计算组成它的素因子的位数和，判断它们是否相等。
AC代码如下：
#include<algorithm> #include<iostream> #include<string.h> #include<cstdio> #include<cstring> #include<cmath> #define ll long long using namespace std; const int maxn=1e5+100; int a[100],b[100],x; bool isPrime(int num) { if (num == 2 || num == 3) { return true; } if (num % 6 != 1 && num % 6 != 5) { return false; } for (int i = 5; i*i <= num; i += 6) { if (num % i == 0 || num % (i+2) == 0) { return false; } } return true; } int get_sum(int x) { int num=0; while(x) { num+=x%10; x/=10; } return num; } int first(int x) { return get_sum(x); } void factor(int n,int a[1000],int b[1000],int &tot) { int temp,i,now; temp=(int)((double)sqrt(n)+1); tot=0; now=n; for(i=2;i<=temp;++i) if(now%i==0){ a[++tot]=i; b[tot]=0; while(now%i==0){ ++b[tot]; now/=i; } } if(now!=1){ a[++tot]=now; b[tot]=1; } } int second(int x) { int tot; memset(a,0,sizeof(a)); memset(b,0,sizeof(b)); factor(x,a,b,tot); int sum=0; for(int i=1;i<=tot;++i) { sum=sum+get_sum(a[i])*b[i]; } return sum; } int main() { while(~scanf("%d",&x)&&x) { int k=1; for(int i=x+1;k==1;++i) { if(isPrime(i)) continue; if(first(i)==second(i)) { printf("%d\n",i); k=0; } } } return 0; }
