约数个数

 

C - 1003

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

Submit Status

 

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.

 

#include<iostream>
#include<cstring>
#include<cmath>
using namespace std;
#define maxn 1000010
#define mod (1<<30)
int prime[maxn];
int p[maxn];
int f=0;
void hehe()
{
	cout<<++f<<endl;
}
int main()
{
	int a,b,c;
	cin>>a>>b>>c;
	int cnt=0;
	for(int i=2;i<=a*b*c;i++)
		if(!p[i])
		{
			prime[++cnt]=i;
			for(int j=i+i;j<=a*b*c;j+=i)
				p[j]=1;
		}
	int ans=1;
	int sum=0;
	for(int i=1;i<=a;i++)
		for(int j=1;j<=b;j++)
			for(int k=1;k<=c;k++)
			{
				ans=1;
				int temp=i*j*k;
				for(int o=1;o<=cnt&&temp>=prime[o];o++)
				{
					if(temp%prime[o]==0)
					{
						int k1=1;
						temp/=prime[o];
						while(temp%prime[o]==0)
						{
							k1++;
							temp/=prime[o];
						}
						ans=(ans*(k1+1))%mod;
					}
				}
				sum=(sum+ans)%mod;
			}
			cout<<sum<<endl;
	return 0;
}