Easy Number Challenge（求因子个数）

 

Crawling in process... Crawling failed Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u

Description

Let's denote d(n) as the number of divisors of a positive integer n. You are given three integers a, b and c. Your task is to calculate the following sum:

Find the sum modulo 1073741824 (2³⁰).

Input

The first line contains three space-separated integers a, b and c (1 ≤ a, b, c ≤ 100).

Output

Print a single integer — the required sum modulo 1073741824 (2³⁰).

Sample Input

Input

2 2 2

Output

20

Input

5 6 7

Output

1520

Sample Output

Hint

For the first example.

• d(1·1·1) = d(1) = 1;
• d(1·1·2) = d(2) = 2;
• d(1·2·1) = d(2) = 2;
• d(1·2·2) = d(4) = 3;
• d(2·1·1) = d(2) = 2;
• d(2·1·2) = d(4) = 3;
• d(2·2·1) = d(4) = 3;
• d(2·2·2) = d(8) = 4.

So the result is 1 + 2 + 2 + 3 + 2 + 3 + 3 + 4 = 20.

题意：即求将每个数的因子个数相加

题解：打表列出每个数的因子个数，相加即可

 

#include   
#define MAXN 1000100  
using namespace std;  
int df[1000500]={0};  
int main()  
{  
     int i,j,k;  
     int a,b,c;  
     int ans=0;  
     for(i=1;i<=MAXN;i++)  
     {  
         df[i]++;  
         j=i+i;  
         while(j<=MAXN)  
         {  
             df[j]++;  
             j+=i;  
         }  
         df[i]%=1073741824;  
     }  
     cin>>a>>b>>c;  
     for(i=1;i<=a;i++)  
         for(j=1;j<=b;j++)  
             for(k=1;k<=c;k++)  
             {  
                 ans+=df[i*j*k];  
                 ans%=1073741824;  
             }  
     cout<