Codeforces 235E Number Challenge (神定理＋莫比乌斯反演)

E. Number Challenge
time limit per test：3 seconds

memory limit per test：512 megabytes

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:



Find the sum modulo 1073741824(230).

Input
The first line contains three space-separated integersa,
b andc (1 ≤ a, b, c ≤ 2000).

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

Sample test(s)

Input

2 2 2

Output

20

Input

4 4 4

Output

328

Input

10 10 10

Output

11536

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.

题目链接：http://codeforces.com/contest/235/problem/E

题目大意：就是算那个公式的值

题目分析：第一次写DIV1的E，还是CLJ出的题，直接给出rng_58给的一个公式吧:


知道这个公式以后基本就可以秒掉这题了，先枚举i的因子，然后在gcd(i, j) = gcd(i, k) = 1的条件下，为了让gcd(j, k) = 1，直接对b,c进行莫比乌斯反演，跑出来2000ms+，这里有个优化，考虑到a，b，c的范围不是很大，可以对gcd记忆化，瞬间变成500ms+

#include <cstdio>
#include <algorithm>
#define ll long long
using namespace std;
int const MAX = 2005;
int const MOD = 1 << 30;
int gd[MAX][MAX], mob[MAX], p[MAX];
bool noprime[MAX];

void Mobius()
{
    int pnum = 0;
    mob[1] = 1;
    for(int i = 2; i < MAX; i++)
    {
        if(!noprime[i])
        {
            p[pnum ++] = i;
            mob[i] = -1;
        }
        for(int j = 0; j < pnum && i * p[j] < MAX; j++)
        {
            noprime[i * p[j]] = true;
            if(i % p[j] == 0)
            {
                mob[i * p[j]] = 0;
                break;
            }
            mob[i * p[j]] = -mob[i];
        }
    }
}

int Gcd(int a, int b)
{
    if(b == 0)
        return a;
    if(gd[a][b])
        return gd[a][b];
    return gd[a][b] = Gcd(b, a % b);
}

ll cal(int d, int x)
{
    ll ans = 0;
    for(int i = 1; i <= d; i++)
        if(Gcd(i, x) == 1)
            ans += (ll) (d / i);
    return ans;
}

int main()
{
    Mobius();
    int a, b, c;
    ll ans = 0;
    scanf("%d %d %d", &a, &b, &c);
    for(int i = 1; i <= a; i++)
        for(int j = 1; j <= min(b, c); j++)
            if(Gcd(i, j) == 1)
                ans = (ans % MOD + (ll) (a / i) * mob[j] * cal(b / j, i) * cal(c / j, i) % MOD) % MOD;
    printf("%I64d\n", ans);
}