HDU 4432(Sum of divisors) 2012 ACM/ICPC 天津赛区现场赛B题-优快云博客

本文介绍了如何解决HDU在线评测平台的4432号题目，即计算不超过100的数的所有因子的平方和，并在十六进制下输出结果。通过提供输入样例和输出样例，详细解释了解题思路和代码实现。

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题目链接地址：http://acm.hdu.edu.cn/showproblem.php?pid=4432

Sum of divisors

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 137 Accepted Submission(s): 54

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

Recommend

zhoujiaqi2010

需要注意的是：在找因子的时候循环从1-->sqrt(n),每次找到一个约数i,则意味着n/i也是n的约数

#include<iostream>
#include<cmath>
#include<stack>
using namespace std;
__int64 n;
int m;
int solve(int n) //求n的m进制中每个数的平方和
{
    int t,num = 0;
    while(n)
    {
        t = n%m;
        num += t*t;
        n /= m;
    }
    return num;
}
void shi_change(__int64 n) //求n的m进制并输出
{
    stack<int> s;
    int t;
    while(n)
    {
        s.push(n%m);
        n /= m;
    }
    while(!s.empty())
    {
        if(s.top() >= 10) printf("%c",'A'+s.top()-10);
        else printf("%d",s.top());
        s.pop();
    }
    printf("\n");
}
int main()
{
    //freopen("4432.txt","r",stdin);
    while(scanf("%I64d%d",&n,&m) != EOF)
    {
        int i,t,ans = 0;
        t = sqrt(n*1.0);
        for(i = 1; i <= t; i++)
            if(n%i == 0)
            {
                ans += solve(i);
                if(i != n/i) ans += solve(n/i);
            }
        shi_change(ans);
    }
    return 0;
}