HDU4432 Sum of Divisors

涉及知识点：

　　1. 进制转换。

　　2. 找因子时注意可以降低复杂度。

 

Sum of divisors

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4837    Accepted Submission(s): 1589

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.

 

Source

2012 Asia Tianjin Regional Contest

 

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#include<stdio.h>
#include<math.h>

int Out[64];

int main() {
    int n, m;
    while(~scanf("%d%d", &n, &m)) {
        int sum = 0;
        int limit = (int)sqrt(n);  // 当i是因子时 n/i通常也是因子 可以将复杂度将为O（logn）。
        for(int i = 1; i <= limit; i++) {
            if(n % i == 0) {
                int t;
                t = i;
                while(t) {
                    sum += (t % m)*(t % m);
                    t /= m;
                }
                if(i * i != n) {  //避免重复计算。
                    t = n/i;
                    while(t) {
                        sum += (t % m) * (t % m);
                        t /= m;
                    }
                }
            }
        }
        int i = 0;
        while(sum) {
            Out[i++] = sum % m;
            sum /= m;
        }
        for(int j = i - 1; j >= 0; j--) {
            if(Out[j] > 9) {
                printf("%c", Out[j] - 10 + 'A');
            } else printf("%d", Out[j]);
        }
        puts("");
    }
    return 0;
}