CF893E. Counting Arrays

E. Counting Arrays

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given two positive integer numbers x and y. An array F is called an y-factorization of x iff the following conditions are met:

• There are y elements in F, and all of them are integer numbers;
• .

You have to count the number of pairwise distinct arrays that are y-factorizations of x. Two arrays A and B are considered different iff there exists at least one index i (1 ≤ i ≤ y) such that Ai ≠ Bi. Since the answer can be very large, print it modulo 109 + 7.

Input

The first line contains one integer q (1 ≤ q ≤ 105) — the number of testcases to solve.

Then q lines follow, each containing two integers xi and yi (1 ≤ xi, yi ≤ 106). Each of these lines represents a testcase.

Output

Print q integers. i-th integer has to be equal to the number of yi-factorizations of xi modulo 109 + 7.

 

dp求每个数可以有多少因子组合。再排列组合位置，和正负号。效率不高，以后再优化。

import java.util.*;
import java.io.*;

public class Main {
	public static void main(String args[]) {new Main().run();}

	FastReader in = new FastReader();
	PrintWriter out = new PrintWriter(System.out);
	void run(){
		work();
		out.flush();
	}
	long mod=1000000007;
	long gcd(long a,long b) {
		return a==0?b:gcd(b%a,a);
	}
	long[] rec;
	long[] rec2;
	long[] pow2;
	void work() {
		rec=new long[1000001];
		rec2=new long[1000001];
		pow2=new long[1000001];
		rec[0]=1;
		rec2[0]=1;
		pow2[0]=1;
		for(int i=1;i<1000001;i++) {
			rec[i]=(rec[i-1]*i)%mod;
			rec2[i]=pow(rec[i],mod-2)%mod;
			pow2[i]=(pow2[i-1]*2)%mod;
		}
		long[][] dp=new long[21][1000001];
		dp[0][1]=1;
		for(int i=0;i<20;i++) {
			for(int j=1;j<1000001;j++) {
				if(dp[i][j]==0)continue;
				for(int k=2;k*j<1000001;k++) {
					dp[i+1][k*j]+=dp[i][j];
					dp[i+1][k*j]%=mod;
				}
			}
		}
		int q=ni();
		for(int i=0;i<q;i++) {
			int x=ni(),y=ni();
			long ret=0;
			for(int j=0;j<=Math.min(20, y);j++) {
				if(dp[j][x]>0) {
					ret+=dp[j][x]*C(j,y);
					ret%=mod;
				}
			}
			ret*=pow2[y-1];
			ret%=mod;
			out.println(ret);
		}
	}
	

	private long C(int j, int y) {
//		long a=rec[y]*pow((rec[j]*rec[y-j])%mod,mod-2);
		long a=((rec[y]*rec2[j])%mod*rec2[y-j])%mod;
		return a%mod;
	}


	private long pow(long a, long c) {
		long r=1;
		while(c>0) {
			if(c%2==1) {
				r*=a;
				r%=mod;
			}
			c/=2;
			a*=a;
			a%=mod;
		}
		return r;
	}


	//input
	private ArrayList<Integer>[] ng(int n, int m) {
		ArrayList<Integer>[] graph=(ArrayList<Integer>[])new ArrayList[n];
		for(int i=0;i<n;i++) {
			graph[i]=new ArrayList<>();
		}
		for(int i=1;i<=m;i++) {
			int s=in.nextInt()-1,e=in.nextInt()-1;
			graph[s].add(e);
			graph[e].add(s);
		}
		return graph;
	}
	
	private ArrayList<long[]>[] ngw(int n, int m) {
		ArrayList<long[]>[] graph=(ArrayList<long[]>[])new ArrayList[n];
		for(int i=0;i<n;i++) {
			graph[i]=new ArrayList<>();
		}
		for(int i=1;i<=m;i++) {
			long s=in.nextLong()-1,e=in.nextLong()-1,w=in.nextLong();
			graph[(int)s].add(new long[] {e,w});
			graph[(int)e].add(new long[] {s,w});
		}
		return graph;
	}

	private int ni() {
		return in.nextInt();
	}

	private long nl() {
		return in.nextLong();
	}

	private long[] na(int n) {
		long[] A=new long[n];
		for(int i=0;i<n;i++) {
			A[i]=in.nextLong();
		}
		return A;
	}
	private int[] nia(int n) {
		int[] A=new int[n];
		for(int i=0;i<n;i++) {
			A[i]=in.nextInt();
		}
		return A;
	}
}	

class FastReader
{
	BufferedReader br;
	StringTokenizer st;

	public FastReader()
	{
		br=new BufferedReader(new InputStreamReader(System.in));
	}


	public String next() 
	{
		while(st==null || !st.hasMoreElements())//回车，空行情况
		{
			try {
				st = new StringTokenizer(br.readLine());
			} catch (IOException e) {
				e.printStackTrace();
			}
		}
		return st.nextToken();
	}

	public int nextInt() 
	{
		return Integer.parseInt(next());
	}

	public long nextLong()
	{
		return Long.parseLong(next());
	}
}