Codeforces #319(Div.2) A. Multiplication Table_codeforces#319 div2 a-优快云博客

本文介绍了一种算法，用于在一个n*n的乘法表中查找指定数值x出现的次数。通过遍历每一行并检查x是否能被整除且商不超过n来确定x出现的位置。

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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 test(s)

input
10 5

output
2

input
6 12

output
4

input
5 13

output
0

可以看到x在每行最多出现一次，在x/i的地方，枚举i判断x%i==0和x/i<=n，更新答案。O(n)

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>

using namespace std;

int main()
{
	int n, x;
	while (scanf("%d%d", &n, &x) != EOF)
	{
		int ans = 0;

		for (int i = 1; i <= n; i++)
		{
			if (!(x%i) && (x / i) <= n)
				ans++;
		}

		printf("%d\n", ans);
	}
}