Codeforces Codeforces Round #319 (Div. 2) A. Multiplication Table 水题

A. Multiplication Table

Time Limit: 1 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/577/problem/A

Description

Let's consider a table consisting of n rows and n columns. The cell located at the intersection of i-th row and j-th column contains number i × j. The rows and columns are numbered starting from 1.

You are given a positive integer x. Your task is to count the number of cells in a table that contain number x.

Input

The single line contains numbers n and x (1 ≤ n ≤ 105, 1 ≤ x ≤ 109) — the size of the table and the number that we are looking for in the table.

Output

Print a single number: the number of times x occurs in the table.

Sample Input

10 5

Sample Output

2

HINT

题意

给你n,m，相当于给了你一个n*n的乘法表，然后问你这个乘法表里面有多少个格子等于m

题解：

sqrt暴力枚举因子就好了

代码：

//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <bitset>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 200500
#define mod 1001
#define eps 1e-9
#define pi 3.1415926
int Num;
//const int inf=0x7fffffff;
const ll inf=;
inline ll read()
{
    ll x=,f=;char ch=getchar();
    while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
    while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
    return x*f;
}
//*************************************************************************************

int main()
{
    int n=read(),x=read();
    int ans=;
    for(int i=;i<=sqrt(x);i++)
    {
        if(x%i==)
        {
            if(i<=n&&x/i<=n)
                ans+=;
            if(i<=n&&i==x/i)
                ans--;
        }
    }
    cout<<ans<<endl;
}