HDOJ4432 Sum of divisors(暴力)

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 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.
 

题意: 给两个数n和m，对于n的每一个约数k，求k在m进制下的每一个数字的平方和， 再将所有平方和相加起来，用m进制输出。

题目怎么说就怎么做，简单粗暴。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<string>
#include<stack>
using namespace std;
#define rep(i,f,t) for(int i = (f), _end = (t); i <= _end; ++i)
typedef long long int64;
#define debug(x) cout<<"debug   "<<x<<endl;
const int maxn = 1000000000+5;
int n,m;
string jz = "0123456789ABCDEF";

int64 gao(int i){
    int64 ans = 0;
    while(i){
        int tmp = i%m;
        i /= m;
        ans += tmp*tmp;
    }
    return ans;
}

void out(int64 ans){
    stack<char> s;
    while(ans){
        int tmp = ans%m;
        ans /= m;
        s.push(jz[tmp]);
    }
    while(!s.empty()){
        printf("%c",s.top());
        s.pop();
    }
    printf("
");
    return ;
}
int main(){
    while( scanf("%d%d",&n,&m) == 2){
        int ans = 0;
        double sn = sqrt(n);
        if((int)sn*(int)sn == n)ans = gao(sn);
        for(int i = 1; i < sn; ++i){
            if(n%i)continue;
            ans += gao(i);
            ans += gao(n/i);
        }
        out(ans);
    }
    return 0;
}

 

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