HDU 4746 Mophues (莫比乌斯反演应用)

Mophues

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 327670/327670 K (Java/Others)

Total Submission(s): 980    Accepted Submission(s): 376

Problem Description

As we know, any positive integer C ( C >= 2 ) can be written as the multiply of some prime numbers:

    C = p1×p2× p3× ... × pk

which p1, p2 ... pk are all prime numbers.For example, if C = 24, then:

    24 = 2 × 2 × 2 × 3

    here, p1 = p2 = p3 = 2, p4 = 3, k = 4

Given two integers P and C. if k<=P( k is the number of C's prime factors), we call C a lucky number of P.

Now, XXX needs to count the number of pairs (a, b), which 1<=a<=n , 1<=b<=m, and gcd(a,b) is a lucky number of a given P ( "gcd" means "greatest common divisor").

Please note that we define 1 as lucky number of any non-negative integers because 1 has no prime factor.

 

Input

The first line of input is an integer Q meaning that there are Q test cases.

Then Q lines follow, each line is a test case and each test case contains three non-negative numbers: n, m and P (n, m, P <= 5×10⁵. Q <=5000).

 

Output

For each test case, print the number of pairs (a, b), which 1<=a<=n , 1<=b<=m, and gcd(a,b) is a lucky number of P.

 

Sample Input

2
10 10 0
10 10 1

 

Sample Output

63
93

 

Source

2013 ACM/ICPC Asia Regional Hangzhou Online

 

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4746



题目大意：定义num[i]为将i唯一分解后全部质因子的个数。要求num[gcd(a, b)] <= p的(a,b)的对数，当中1 <= a <= n。1 <= b <= m



题目分析：这题T了一天，先不考虑1，题目给的p的最大值是5e5，因此num[i]最大为18，由于2^19就大于5e5了

定义f(d)为gcd(a,b) = d的个数，g(d)为gcd(a,b) = d的倍数的个数，显然g(d)非常好求。就是(n / d) * (m / d)

又g(d) = f(d) + f(2d) + f(3d) + ...对此式进行莫比乌斯反演得到

f(d) = u(1)g(d) + u(2)g(2d) + u(3)g(3d) + ...。最后答案为Σu(k)g(kd)，此时若直接枚举d，就算计算f(d)用分块求和优化成接近sqrt(n)。n*sqrt(n)接近2e8肯定超时，因此要换别的思路，考虑到num[i]最大仅仅有18。我们能够预处理出以i为最大公约数。且分解i后质因子个数等于num[i]的方案数。依据公式有sum[ki][num[i]] += u[k]。令j＝ki，则sum[j][num[i]] += u[j / i]，注意这里算的仅仅是莫比乌斯函数的贡献值，

举个样例。比方

f(2) = u(1)g(2) + u(2)g(4) + u(3)g(6) + ...

f(3) = u(1)g(3) + u(2)g(6) + u(3)g(9) + ...

答案肯定要把它们加起来，注意到g(6)出现了两次，能够理解为6这个数字对num[i] = 1的情况有两次贡献, 因此能够写成(u(2) + u(3)) * g(6)

然后再处理分解i后质因子个数小于等于num[i]的方案数，最后再处理以i为最大公约数的前缀和，预处理工作就结束了，复杂度为nlogn。在线计算时用分块求和优化，复杂度为qsqrt(n)。总的复杂度大概为nlogn+qsqrt(n)

#include <cstdio>
#include <cstring>
#include <algorithm>
#define ll long long
using namespace std;
int const MAX = 5e5 + 5;
int n, m, p, pnum;
int mob[MAX], pr[MAX], sum[MAX][20];
int num[MAX];
bool prime[MAX];

void Mobius()
{
    pnum = 0;
    memset(prime, true, sizeof(prime));
    mob[1] = 1;
    for(int i = 2; i < MAX; i++)
    {
        if(prime[i])
        {
            pr[pnum ++] = i;
            num[i] = 1;
            mob[i] = -1;
        }
        for(int j = 0; j < pnum && i * pr[j] < MAX; j++)
        {
            num[i * pr[j]] = num[i] + 1;
            prime[i * pr[j]] = false;
            if(i % pr[j] == 0)
            {
                mob[i * pr[j]] = 0;
                break;
            }
            mob[i * pr[j]] = -mob[i];
        }
    }
}

void Init()
{
    Mobius();
    for(int i = 1; i < MAX; i++)
        for(int j = i; j < MAX; j += i)
            sum[j][num[i]] += mob[j / i];
    for(int i = 1; i < MAX; i++)
        for(int j = 1; j < 19; j++)
            sum[i][j] += sum[i][j - 1];
    for(int i = 1; i < MAX; i++)
        for(int j = 0; j < 19; j++)
            sum[i][j] += sum[i - 1][j];
}

ll cal(int l, int r)
{
    ll ans = 0;
    if(l > r)
        swap(l, r);
    for(int i = 1, last = 0; i <= l; i = last + 1)
    {
        last = min(l / (l / i), r / (r / i));
        ans += (ll) (l / i) * (r / i) * (sum[last][p] - sum[i - 1][p]);
    }
    return ans;
}   

int main()
{
    Init();
    int T;
    scanf("%d", &T);
    while(T --)
    {
        scanf("%d %d %d", &n, &m, &p);
        printf("%lld\n", p > 18 ? (ll) n * m : cal(n, m));
    }
}