CodeForces - 236B Easy Number Challenge （暴力）

64bit IO Format: %I64d & %I64u

CodeForces - 236B

Easy Number Challenge

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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

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.

Source

Codeforces Round #146 (Div. 2)

//题意：定义d(n)为n的因子个数

现在输入三个整数a,b,c,让求

思路：

直接暴力加判断

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<iostream>
#define ll long long
#define N 1000010
#define M 1073741824
using namespace std;
int dp[N];
int vis[N];
int main()
{
	int a,b,c;
	int i,j,k;
	while(scanf("%d%d%d",&a,&b,&c)!=EOF)
	{
		ll sum=0;
		memset(vis,0,sizeof(vis));
		memset(dp,0,sizeof(dp));
		for(i=1;i<=a;i++)
		{
			for(j=1;j<=b;j++)
			{
				for(k=1;k<=c;k++)
				{
					int m=i*j*k;
					int cnt=0;
					if(!vis[m])
					{
						vis[m]=1; 
						for(int ii=1;ii<=sqrt(m);ii++)
						{
							if(m%ii==0)
							{
								if(m/ii==ii)
									cnt+=1;
								else
									cnt+=2; 
							}
						}
						dp[m]=cnt;
					}
					sum=(sum+dp[m])%M;
				}
			}
		}
		printf("%lld\n",sum);
	}
	return 0;
}