HDU - 4746 Mophues

Mophues

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 327670/327670 K (Java/Others)
Total Submission(s): 2824    Accepted Submission(s): 1215

 

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

 

Recommend

liuyiding

 

 

分析：显而易见，当p比较大时，ans一定是n*m，那么也就是log_2^(1e6)=20。

令num[i]表示i的素因子个数。

考虑u(d)的意义，令

u(d)仅在d无平方因子时才有值，此时i/d一定包含了所有d中所有素因子次数-1的因子，那么 ，num[i]同上，num1[i]表示i的素因子种类数。

那么sum(t,p)对p求一个前缀和就是

至此就可以 求解了。

#include <bits/stdc++.h>
using namespace std;
long long num[500004],num1[500004];
long long sum[500004][20];
long long fac[24];
bool vis[500004];
long long c(int a,int b){
    if(b<0)return 0;
    if(a<b)return 0;
    if(a==b)return 1;
    return fac[a]/fac[b]/fac[a-b];
}
vector<int>prim;
int main(){
    int t;
    cin>>t;
    fac[0]=1;
    for (int i = 1; i < 15; ++i) {
        fac[i]=fac[i-1]*i;
    }
    for (int i = 2; i < 500004; ++i) {
        if(!vis[i]){
            prim.push_back(i);
            num[i]=1;
        }
        for (int j = 0; j < prim.size() && i*prim[j] < 500004; ++j) {
            vis[i*prim[j]]=1;
            num[i*prim[j]] = num[i]+1;
            if(i%prim[j] == 0)break;
        }
    }

    for (int i = 0; i < prim.size(); ++i) {
        for (int j = prim[i]; j < 500004; j+=prim[i]) {
            num1[j]++;
        }
    }
    for (int i = 1; i <= 500004; ++i) {
        for (int j = 0; j < 20; ++j) {
            sum[i][j] = c(num1[i],j - num[i] + num1[i]);
            if((num[i]-j)&1)sum[i][j]*=-1;
            if(j)sum[i][j]+=sum[i][j-1];
        }
    }
    for (int j = 0; j < 20; ++j) {
        for (int i = 1; i <= 500004; ++i) {
            sum[i][j] += sum[i-1][j];
        }
    }
    while(t--){
        int n,m,p;
        scanf("%d%d%d",&n,&m,&p);
        if(p >= 20){
            printf("%lld\n",1LL*n*m);
            continue;
        }
        if(n>m)swap(n,m);
        long long ans = 0;
        for (int l = 1,r; l <= n; l=r+1) {
            r = min(n/(n/l),m/(m/l));
            if (r >= n)r=n;
            ans += 1LL*(n/l)*(m/l)*(sum[r][p] - sum[l-1][p]);
        }
        cout<<ans<<endl;
    }
}