B. Easy Number Challenge

最新推荐文章于 2020-04-13 14:48:00 发布

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本文介绍了一个关于计算特定范围内正整数除数数量总和的问题，并提供了一种通过预处理和查找因子来求解的方法。具体而言，任务是计算给定三个整数a、b、c的情况下所有形如i*j*k (1 ≤ i ≤ a, 1 ≤ j ≤ b, 1 ≤ k ≤ c)的数的除数个数之和。

题目链接:传送门

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

$\text{[math]}$

Find the sum modulo 1073741824(2³⁰).

Input

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

Output

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

Sample test(s)

Input

2 2 2

Output

Input

5 6 7

Output

Note

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.

意解:暴力,先把暴力枚举一边,记录每个数出现的个数,然后Nsqrt(N)找因子;

AC代码:

#include <iostream>
#include <vector>
#include <cstdio>
#include <cmath>
#define For(i,a,n) for(int i = a; i <= n; i++)

using namespace std;

const int M = 1e6;
const int INF = 1LL << 30;
int ma[M + 10];

int main()
{
    int a,b,c;
    int ans = 0;
    scanf("%d %d %d",&a,&b,&c);
    For(i,1,a)
     For(j,1,b)
      For(k,1,c)
           ma[i * j * k]++;
    for(int i = 1; i <= M; i++)
    {
        if(ma[i])
        for(int p = 1; p * p <= i; p++)
           {
               if(i % p == 0)
               {
                   ans += ma[i];
                   if(p * p != i) ans += ma[i];
               }
           }
    }
    cout<<ans<<endl;
    return 0;
}