hdu 4432 因子

Sum of divisors

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5052    Accepted Submission(s): 1673

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

Output the answer base m.

 

Sample Input

10 230 5

 

Sample Output

110112
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

区域赛水题

取因子

这里用sqrt （n）    同时取两个因子，不需要标记

      若一个是  i   另外一个就是   n / i   ，因为是开根号就不用标记，不用怕之前已经被用过了

#include<stdio.h>#include<math.h>#include<string.h>#include<algorithm>using namespace std;long long n;int m;int a[100000000];long long pos;int main(){    //freopen("in.txt","r",stdin);    while(scanf("%I64d%d",&n,&m)!=EOF)    {        long long ans=0;        long long number=sqrt(n*1.0);        long long x,i;        for(i=1;i<=number;i++){            if(n%i==0){                x=i;                while(x)                {                    int t=x%m;                    ans+=t*t;                    x=x/m;                }                x=n/i;                while(x)                {                    int t=x%m;                    ans+=t*t;                    x=x/m;                }            }        }        if(n==1)            ans=1;        pos=0;        while(ans)        {            a[pos++]=ans%m;            ans/=m;        }        pos--;        for(long long i=pos;i>=0;i--){            if(a[i]>9)                printf("%c",a[i]+'A'-10);            else                printf("%d",a[i]);        }        puts("");    }    return 0;}