poj1142 Smith Numbers
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
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