Codeforces 235E Number Challenge (神定理＋莫比烏斯反演)

_TCgogogo_發表於2015-08-26



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.


題目連結：http://codeforces.com/contest/235/problem/E


題目大意：就是算那個公式的值


題目分析：第一次寫DIV1的E，還是CLJ出的題，直接給出rng_58給的一個公式吧:

知道這個公式以後基本就可以秒掉這題了，先列舉i的因子，然後在gcd(i, j) = gcd(i, k) = 1的條件下，為了讓gcd(j, k) = 1，直接對b,c進行莫比烏斯反演，跑出來2000ms+，這裡有個優化，考慮到a，b，c的範圍不是很大，可以對gcd記憶化，瞬間變成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);
}


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