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;
}