# 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 :

```                            ```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.IOException;
import java.util.InputMismatchException;
import java.io.PrintStream;
import java.io.OutputStream;
import java.io.PrintWriter;
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);
solver.solve(1, in, out);
out.close();
}
}

/////////////////////////////////////////////////////////////
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);
}
}

public FastScanner(InputStream is) {
}

try {
//            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() {
while (isWhiteSpace(c)) {
}
int sgn = 1;
if (c == '-') {
sgn = -1;
}
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';
}
return ret * sgn;
}

try {
} 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 = 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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