Easy Number Challenge(暴力，求因子个数)

Easy Number Challenge

Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Submit Status Practice CodeForces 236B

Appoint description: 
System Crawler  (2016-04-26)

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.

题解：

d(x)代表x的因子的个数；还好i,j,k都不大，100，暴力就行，直接由于因子个数等于质因子的系数加一之积，反素数讲过，由此可得；

代码：

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<iostream>
#include<set>
#define ll long long
#define MOD 1073741824
using namespace std;
int num[];
int main()
{
    int a,b,c;
    int i,j,k;
    while(scanf("%d%d%d",&a,&b,&c)!=EOF)
    {
        memset(num, , sizeof(num));
        ll sum=, temp;
        set<int>st;
        set<int>::iterator iter;
        for(i=;i<=a;i++)
        {
            for(j=;j<=b;j++)
            {
                for(k=;k<=c;k++)
                {
                temp = i * j * k;
                ll cnt = ;
                for(int p = ; p <= temp; p++){
                    if(temp % p == ){
                        int cur = ;
                        while(temp % p == ){
                            temp /= p;
                            cur++;
                        }
                        cnt *= cur + ;
                    }
                }
                sum += cnt;
                sum %= MOD;
                }
                }
            }

        printf("%lld\n",sum);
    }
    return ;
}