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

210 10 010 10 1

 

Sample Output

6393

 

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唯一分解后所有质因子的个数，要求num[gcd(a, b)] <= p的(a,b)的对数，其中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，因为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)很好求，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]最大只有18，我们可以预处理出以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，根据公式有sum[ki][num[i]] += u[k]，令j＝ki，则sum[j][num[i]] += u[j / i]，注意这里算的只是莫比乌斯函数的贡献值，
举个例子，比如
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，可以理解为6这个数字对num[i] = 1的情况有两次贡献, 因此可以写成(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 longusing 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));    }}