# Sum of divisors

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 82    Accepted Submission(s): 35

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≤109
2≤m≤16
There are less then 10 test cases.

Output

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

Recommend
zhoujiaqi2010

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
#include<math.h>
using namespace std;

int bit;
int cnt;

void change(int n,int base)
{
cnt=0;
while(n)
{
bit[cnt++]=n%base;
n/=base;
}
}
int main()
{
int n,m;
while(scanf("%d%d",&n,&m)!=EOF)
{
int sum=0;
int t=(int)sqrt(n*1.0);
for(int i=1;i<=t;i++)
{
if(n%i==0)
{
int tmp=i;
while(tmp)
{
sum+=((tmp%m)*(tmp%m));
tmp/=m;
}
tmp=n/i;
if(tmp==i)continue;
while(tmp)
{
sum+=((tmp%m)*(tmp%m));
tmp/=m;
}
}
}
change(sum,m);
for(int i=cnt-1;i>=0;i--)
{
if(bit[i]>9)printf("%c",bit[i]-10+'A');
else printf("%d",bit[i]);
}
printf("\n");
}
return 0;
}

posted on 2012-10-26 14:23  kuangbin  阅读(...)  评论(...编辑  收藏

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