P-sequences


Problem Statement :


We call a sequence of N natural numbers (a1, a2, ..., aN) a P-sequence, if the product of any two adjacent numbers in it is not greater than P. In other words, if a sequence (a1, a2, ..., aN) is a P-sequence, then ai * ai+1 ≤ P ∀ 1 ≤ i < N

You are given N and P. Your task is to find the number of such P-sequences of N integers modulo 109+7.

Input Format

The first line of input consists of N
The second line of the input consists of P.

Constraints

2 ≤ N ≤ 103
1 ≤ P ≤ 109
1 ≤ ai

Output Format

Output the number of P-sequences of N integers modulo 109+7.


Solution :



title-img


                            Solution in C :

In C++ :





#include<math.h>
#include<algorithm>
#include<cstdlib>
#include<iostream>
#include<stdio.h>
#include<map>
#include<ext/hash_map>
#include<ext/hash_set>
#include<set>
#include<string>
#include<assert.h>
#include<vector>
#include<time.h>
#include<queue>
#include<deque>
#include<sstream>
#include<stack>
#include<sstream>
#define MA(a,b) ((a)>(b)?(a):(b))
#define MI(a,b) ((a)<(b)?(a):(b))
#define AB(a) (-(a)<(a)?(a):-(a))
#define X first
#define Y second
#define mp make_pair
#define pb push_back
#define pob pop_back
#define ep 0.0000000001
#define Pi 3.1415926535897932384626433832795
using namespace std;
using namespace __gnu_cxx;
const long long  MO=1000000000+7;
int x,y,i,j,k,n,m,mid,l,r;
int a[100000],p,kk;
long long b[2][100000];
int main()
{
    cin>>n>>p;
    kk=k=int(sqrt(p));
    for (i=1;i<=k;i++) a[i]=i;
    for (i=kk;i>=1;i--)
    {
        if (p/i>a[k]) {k++; a[k]=p/i;}
    }
    for (i=1;i<=k;i++)
    b[1][i]=a[i];
    b[0][0]=0;
    b[1][0]=0;
    for (i=2;i<=n;i++)
    {
        r=k;
        for (j=1;j<=k;j++){
            while (a[r]*a[j]>p) r--;
            b[(i&1)][j]=(b[(!(i&1))][r]*(a[j]-a[j-1])+b[(i&1)][j-1])%MO;
   //     cout<<b[(i&1)][j]<<" "<<a[j]<<" -- ";
        }
   //     cout<<endl;
    }
    cout<<b[(n&1)][k]<<endl;

    return 0;
}








In Java :





import java.io.InputStreamReader;
import java.io.IOException;
import java.util.InputMismatchException;
import java.io.PrintStream;
import java.io.BufferedReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.io.Reader;
import java.io.Writer;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 * @author Nipuna Samarasekara
 */
public class Solution {
	public static void main(String[] args) {
		InputStream inputStream = System.in;
		OutputStream outputStream = System.out;
		FastScanner in = new FastScanner(inputStream);
		FastPrinter out = new FastPrinter(outputStream);
		Task3 solver = new Task3();
		solver.solve(1, in, out);
		out.close();
	}
}

class Task3 {
    /////////////////////////////////////////////////////////////
  static long mod=1000000007;
    public void solve(int testNumber, FastScanner in, FastPrinter out) {
   long N=in.nextInt(),p=in.nextInt();
   int sqrt= (int)Math.sqrt(p) ;
     long[][] finishes= new long[2][1+sqrt];
     long[][] finpdivupw= new long[2][1+sqrt];
        for (int i = 1; i <=sqrt ; i++) {
          finishes[0][i]=1;
          if(i<sqrt)
          finpdivupw[0][i]= p/i - p/(i+1);
        }
        finpdivupw[0][sqrt]=p/sqrt-sqrt;
       int cur=0,prev=1;
        for (int i = 0; i < N-1; i++) {
          cur^=1;
          prev^=1;
            long sum=0;
            for (int j = 1; j < sqrt ; j++) {
                sum+=finishes[prev][j];
                if(sum>=mod)sum%=mod;
                long val=sum*(p/j - p/(j+1));
                val%=mod;
                finpdivupw[cur][j]=val;

            }
            sum+=finishes[prev][sqrt];
            long val=  sum*(p/sqrt-sqrt);
            finpdivupw[cur][sqrt]= val%mod;
            for (int j = sqrt; j >= 1; j--) {
              sum+=finpdivupw[prev][j];
                if(sum>=mod)sum%=mod;
                finishes[cur][j]=sum;
            }



        }
        long ans=0;
        for (int i = 1; i <= sqrt ; i++) {
           ans+=finishes[cur][i];
            if(ans>=mod)ans%=mod;
           ans+=finpdivupw[cur][i];
            if(ans>=mod)ans%=mod;
        }
        out.println(ans);
    }
}

class FastScanner extends BufferedReader {

    public FastScanner(InputStream is) {
        super(new InputStreamReader(is));
    }

    public int read() {
        try {
            int ret = super.read();
//            if (isEOF && ret < 0) {
//                throw new InputMismatchException();
//            }
//            isEOF = ret == -1;
            return ret;
        } catch (IOException e) {
            throw new InputMismatchException();
        }
    }

    static boolean isWhiteSpace(int c) {
        return c >= 0 && c <= 32;
    }

    public int nextInt() {
        int c = read();
        while (isWhiteSpace(c)) {
            c = read();
        }
        int sgn = 1;
        if (c == '-') {
            sgn = -1;
            c = read();
        }
        int ret = 0;
        while (c >= 0 && !isWhiteSpace(c)) {
            if (c < '0' || c > '9') {
                throw new NumberFormatException("digit expected " + (char) c
                        + " found");
            }
            ret = ret * 10 + c - '0';
            c = read();
        }
        return ret * sgn;
    }

    public String readLine() {
        try {
            return super.readLine();
        } catch (IOException e) {
            return null;
        }
    }

}

class FastPrinter extends PrintWriter {

    public FastPrinter(OutputStream out) {
        super(out);
    }

    public FastPrinter(Writer out) {
        super(out);
    }


}








In C :





#include <stdio.h>

#define P 1000000007LL

long long b[100000][3],i,j,k,l,m,n,t;
long long a[1010][65000];

int main()
{

scanf("%lld %lld",&n,&t);

k = 1;
l = 0;

while(k<=t)
 {
   b[l][0] = k;
   b[l][1] = t/(t/k);
   b[l][2] = b[l][1] - b[l][0] + 1;
   k = b[l][1] + 1;
   l++;
 }

for(i=0;i<l;i++) a[0][i] = 0;
a[0][0] = 1;

for(i=1;i<=n;i++)
{
 k = 0;
 for(j=0;j<l;j++)
   {
    k = (k+a[i-1][j]*b[j][2])%P;
    a[i][l-1-j] = k;   
   }
}

k = 0;

for(i=0;i<l;i++) k = (k+a[n][i]*b[i][2])%P;

printf("%lld\n",k);

return 0;
}








In Python3 :





#!/bin/python3

import os
import sys

#
# Complete the pSequences function below.
#
def pSequences(n, p):

    MOD = 10**9+7

    Pf = set()
    for i in range(1,int(p**0.5)+1) :
        Pf.add(i)
        Pf.add(p//i)
    Pf = sorted(Pf)
    # print(Pf)

    Ps = []
    for f in Pf :
        Ps.append(f)

    Pfd = [0]
    for i in range(1, len(Pf)) :
        Pfd.append(Pf[i]-Pf[i-1])

    # print(Pfd)
    for n in range(1,n):
    #    print(Ps)
        Ps_ = []
        Ps_.append(Ps[-1])
        for i in range(1,len(Pf)) :
            Ps_.append((Ps_[-1] + Pfd[i] * Ps[-(i+1)]) % MOD)
        Ps = Ps_
            
    # print(Ps)
    return(Ps[-1])

if __name__ == '__main__':
    fptr = open(os.environ['OUTPUT_PATH'], 'w')

    n = int(input())

    p = int(input())

    result = pSequences(n, p)

    fptr.write(str(result) + '\n')

    fptr.close()
                        




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