hdu-4432-Sum of divisors

/*
Sum of divisors

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1996    Accepted Submission(s): 679


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.
 
*/
#include 
#include
#include
#include
#include
#include<string.h>
#include
#include
using namespace std;
#define maxn 26000
int a[maxn];
int sum;
int k,i,j,tmp;
char b[maxn];
void yinzi(int n)
{
    k=0;
    for(i=1; i*i<=n; i++)
    {
        if(n%i==0)
        {
            a[k++]=i;
            if(i!=n/i)
                a[k++]=n/i;
        }
    }
}
void change(int m)
{
    sum=0;
    for(i=0; i)
    {
        while(a[i])
        {
            tmp=a[i]%m;
            sum+=tmp*tmp;
            a[i]=a[i]/m;
        }
    }
}
void rechange(int x,int m)
{
    j=0;
    stack<char> st;
    while(x)
    {
        tmp=x%m;
        if(tmp>=10)
        st.push('A'+tmp-10);
            //b[j++]='A'+tmp-10;
        else
        st.push(tmp+'0');
           // b[j++]=tmp+'0';
        x=x/m;
    }
    while(!st.empty())
    {
        cout<<st.top();
        st.pop();
    }
    /*if(x)
    {
        rechange(x/m,m);
        tmp=x%m;
        if(tmp>=10)
        printf("%c",'A'+tmp-10);
        else
        printf("%d",tmp);
    }*/
}
int main()
{
    int n,m;
    while(~scanf("%d%d",&n,&m))
    {

        yinzi(n);
        change(m);
        rechange(sum,m);
        //cout<<" j= "<
        /*for(i=j-1;i>=0;i--)
        {
            //printf("%c",b[i]);
            cout<*/
        printf("\n");
    }
    return 0;
}

#include 
#include 
#include <string.h>
#include 
using namespace std;
int cal;
void Base(int n,int m){
    if(n)
     {
        Base(n/m,m);
        cal+=(n%m)*(n%m);
     }
}
void out(int n,int m){
    if(n){
        out(n/m,m);
        if(n%m>9)
            printf("%c",'A'+(n%m)-10);
        else printf("%d",n%m);
    }
}
int main(){

  int n,m;
  int i,k,sum;
  while(scanf("%d %d",&n,&m)!=EOF){
      //k=sqrt(n+1.0);
      sum=0;
      for(i=1;i*i)
      {
          if(n%i==0){
            cal=0;
            Base(i,m);
            sum+=cal;
            cal=0;
            Base(n/i,m);
            sum+=cal;
          }
      }
      if(i*i==n){
            cal=0;
            Base(i,m);
            sum+=cal;
      }
      out(sum,m);
      printf("\n");
  }
  return 0;
}

 

/*错误代码，求解*/
#include 
#include
#include
#include
#include
#include<string.h>
#include
#include
using namespace std;
#define maxn 260000
int a[maxn];
int sum;
int k,i,tmp;
char b[maxn];
void yinzi(int n)
{
    k=0;
    for(i=1;i*i<=n;i++)
    {
        if(n%i==0)
        {
            a[k++]=i;
            if(i!=n/i)
                a[k++]=n/i;
        }
    }
}
void change(int m)
{
    sum=0;
    for(i=0;i)
    {
        while(a[i])
        {
            tmp=a[i]%m;
            sum+=tmp*tmp;
            a[i]=a[i]/m;
        }
    }
}

int main()
{
    int n,m,x,j;
    while(~scanf("%d%d",&n,&m))
    {
        yinzi(n);
        change(m);
        // rechange(sum,m);
        x=sum;
        j=0;
        while(x)
        {
            tmp=x%m;
            if(tmp>=10)
                b[j++]=tmp-10+'A';
            else
                b[j++]=tmp+'0';
            x=x/m;
        }
        for(i=j-1;i>=0;i--)
        {
            printf("%c",b[i]);
        }
        printf("\n");
    }
    return 0;
}

 

转载于:https://www.cnblogs.com/heqinghui/p/3233143.html