[hdu4432]Sum of divisors

Sum of divisors

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2960    Accepted Submission(s): 1044

Problem Description

mmm is learning division, she's so proud of herself that she can figure out the sum of all the divisors of numbers no larger than 100 within one day!
But her teacher said "What if I ask you to give not only the sum but the square-sums of all the divisors of numbers within hexadecimal number 100?" mmm get stuck and she's asking for your help.
Attention, because mmm has misunderstood teacher's words, you have to solve a problem that is a little bit different.
Here's the problem, given n, you are to calculate the square sums of the digits of all the divisors of n, under the base m.

 

Input

Multiple test cases, each test cases is one line with two integers.
n and m.(n, m would be given in 10-based)
1≤n≤10 ⁹
2≤m≤16
There are less then 10 test cases.

 

Output

Output the answer base m.

 

Sample Input

   
   
   
   

    
    
    
    10 2
30 5
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    110
112

    
    
    
    
 
     
 
      Hint 
     
 Use A, B, C...... for 10, 11, 12...... Test case 1: divisors are 1, 2, 5, 10 which means 1, 10, 101, 1010 under base 2, the square sum of digits is 1^2+ (1^2 + 0^2) + (1^2 + 0^2 + 1^2) + .... = 6 = 110 under base 2. 
    
    
    
    
     
   
   
   
   

 

Source

2012 Asia Tianjin Regional Contest


题目本身没有什么好说的 注意枚举约数的时候枚举到sqrt(n)就行了
每次处理了当前约数的时候不要忘了还有n/(当前约数) 这个约数
还有最后输出答案时如果进制大于10的时候要转换成字母 之前用了很蠢的方法....

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
int main()
{
    int n,m;
    while (scanf("%d%d",&n,&m)==2)
    {
        int ans=0;
        for (int i=1;i<=(int)sqrt(n);i++)
        {
            if(n%i==0)
            {
                int t=i;
                while(t>0)
                {
                    ans+=((t%m)*(t%m));
                    t/=m;
                }
                t=n/i;
                if (t==i) continue;
                while(t>0)
                {
                    ans+=((t%m)*(t%m));
                    t/=m;
                }
            }
        }
        int b[1000],cnt=0;
        while (ans>0)
        {
            b[cnt++]=ans%m;
            ans/=m;
        }
        for (int i=cnt-1;i>=0;i--)
        {
            if (b[i]>9) printf("%c",b[i]-10+'A');
            else printf("%d",b[i]);
        }
        printf("\n");
    }
    return 0;
}