hdu4432 Sum of divisors（数论）

Sum of divisors

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2063    Accepted Submission(s): 718

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

 

现在看觉得挺简单的，第一眼看时觉得挺难就没做，谁知道人家一下就AC了，直接一般的方法做加一点优化就行了

#include <iostream>

#include <stdio.h>

#include <math.h>

using namespace std;

int vet[10000];

int main()

{

    int n,m,all,sum,s,ii,i,t;

    char str[17]={'0','1','2','3','4','5','6','7','8','9','A','B','C','D','E'};

    while(scanf("%d%d",&n,&m)!=EOF)

    {

        if(n==1)

        {

            printf("1\n");

            continue;

        }

        all=n;sum=0;

        for(i=2;i<all;i++)

        {

            if(n%i==0)

            {

                all=n/i;//注意这一步优化，最关键的一步，如果i是n的因数，那么我们直接把i和n/i一起算了，i的上界变成了n/i-1

               // printf("%d %d",i,n/i);

                s=0;

                ii=i;

                while(ii)

                {

                    t=ii%m;

                    s+=t*t;

                    ii=ii/m;

                }

                ii=n/i;

                while(ii)

                {

                    t=ii%m;

                    s+=t*t;

                    ii=ii/m;

                }

                sum+=s;

            }

        }

         s=0;

        ii=n;

        while(ii)

        {

            t=ii%m;

            s+=t*t;

            ii=ii/m;

        }

        sum+=s;

        sum++;

        t=0;

        while(sum)

        {

           vet[t++]=sum%m;

           sum=sum/m;

        }

        for(i=t-1;i>=0;i--)

        {

            printf("%c",str[vet[i]]);

        }



        printf("\n");



    }

    return 0;

}