hdu 4746Mophues[莫比乌斯反演]

Mophues

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 327670/327670 K (Java/Others)
Total Submission(s): 1669    Accepted Submission(s): 675
　

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

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//Source：http://acm.hdu.edu.cn/showproblem.php?pid=4746

Description(题意):

任何整数C
( C >= 2 )都可以写成素数之积
C = p1×p2×
p3×
... × pk
其中， p1, p2 ... pk 是素数。如
C = 24, 则
24 = 2 ×
2 × 2
× 3,
其中, p1 = p2 = p3 = 2, p4 = 3, k = 4.
给定两整数 P和 C,
若 k<=P ( k是
C的素因子个数),称
C是P的幸运数.
现小X需计算的点对 (a,
b)的个数,其中1<=a<=n
, 1<=b<=m, gcd(a,b)是 P的幸运数
( “gcd”是最大公因数).
注意：因为1无素因子，定义1为任何非负数的幸运数.

Input

首行有一个整数
T，表示有 T 组测试数据.接下来有T行，每行是一种测试数据，含3个非负整数n,
m 与P (n, m, P <= 5×105.
T <=5000).

Output

对每种测试数据，输出对
(a, b)的个数，其中 1<=a<=n , 1<=b<=m,
且 gcd(a,b)
是 P的幸运数.

Sample
Input

2

10 10 0

10 10 1

Sample
Output

63

93

//num[j]记录j的因子数。
//g[j][num[i]]用于计算具有相同个数的素因子的i的?(j/i)之和，
#include<cstdio>
#include<iostream>
using namespace std;
typedef long long ll;
const int M=5e5+,N=;
int n,m,p,T,g[M][N],num[M];
int tot,prime[M/],mu[M];bool check[M];
int calc(int y,int x){
    int res=;
    while(!(y%x)) y/=x,res++;
    return res;
}
void sieve(){
    n=5e5;mu[]=;
    for(int i=;i<=n;i++){
        if(!check[i]) prime[++tot]=i,mu[i]=-;
        for(int j=;j<=tot&&i*prime[j]<=n;j++){
            check[i*prime[j]]=;
            if(!(i%prime[j])){mu[i*prime[j]]=;break;}
            else mu[i*prime[j]]=-mu[i];
        }
    }
    for(int i=;i<=n;i++) if(!num[i]) for(int j=i;j<=n;j+=i) num[j]+=calc(j,i);
    for(int i=;i<=n;i++) for(int j=i;j<=n;j+=i) g[j][num[i]]+=mu[j/i];
    for(int i=;i<=n;i++) for(int j=;j<;j++) g[i][j]+=g[i][j-];
    for(int i=;i<=n;i++) for(int j=;j<;j++) g[i][j]+=g[i-][j];
}
ll solve(int n,int m,int p){
    if(p>=) return 1LL*n*m;
    if(n>m) swap(n,m);
    ll ans=;
    for(int i=,pos=;i<=n;i=pos+){
        pos=min(n/(n/i),m/(m/i));
        ans+=1LL*(n/i)*(m/i)*(g[pos][p]-g[i-][p]);
    }
    return ans;
}
int main(){
    sieve();
    for(scanf("%d",&T);T--;){
        scanf("%d%d%d",&n,&m,&p),
        printf("%I64d\n",solve(n,m,p));
    }
    return ;
}