Day 45 D. Buying Shovels

最新推荐文章于 2021-09-01 23:39:45 发布

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最新推荐文章于 2021-09-01 23:39:45 发布

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分类专栏： Codeforces 文章标签： c++ 算法线性代数

本文链接：https://blog.youkuaiyun.com/Jackey_tao/article/details/119962443

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该博客主要讨论了一个数学问题，即如何确定购买特定数量铲子所需的最少包裹数。问题中，商店提供不同类型的铲子包裹，每种类型包含一定数量的铲子。博主通过编程解决了一个输入案例，例如当需要87把铲子且有81种包裹类型时，答案是8。博主还提供了一个解决方案，用于找出当总铲子数和包裹类型数给定时的最小包裹数。

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Problem
Polycarp wants to buy exactly n shovels. The shop sells packages with shovels. The store has k types of packages: the package of the i-th type consists of exactly i shovels (1≤i≤k). The store has an infinite number of packages of each type.

Polycarp wants to choose one type of packages and then buy several (one or more) packages of this type. What is the smallest number of packages Polycarp will have to buy to get exactly n shovels?

For example, if n=8 and k=7, then Polycarp will buy 2 packages of 4 shovels.

Help Polycarp find the minimum number of packages that he needs to buy, given that he:

will buy exactly n shovels in total;
the sizes of all packages he will buy are all the same and the number of shovels in each package is an integer from 1 to k, inclusive.

Input
The first line contains an integer t (1≤t≤100) — the number of test cases in the input. Then, t test cases follow, one per line.

Each test case consists of two positive integers n (1≤n≤109) and k (1≤k≤109) — the number of shovels and the number of types of packages.

Output
Print t answers to the test cases. Each answer is a positive integer — the minimum number of packages.

Example
input
5
8 7
8 1
6 10
999999733 999999732
999999733 999999733
output
8
1
999999733
1

Note
The answer to the first test case was explained in the statement.

In the second test case, there is only one way to buy 8 shovels — 8 packages of one shovel.

In the third test case, you need to buy a 1 package of 6 shovels.

#include<iostream>
#include<algorithm>
#include<string.h>
#include<vector>
#include<stdio.h>
#include<limits.h>
#include<cmath>
#include<set>
#include<map>
#define ll long long
const int mod = 1e9 + 7;
const int INF = 0x3f3f3f3f;
using namespace std;
int main() 
{
    int t;
    cin >> t;
    while (t--) 
    {
        int n, k;
        cin >> n >> k;
        if (n <= k) 
        {
            cout << 1 << endl;
        }
        else 
        {
            bool f = 0;
            ll ans = INF;
            for (ll i = 1; i * i <= n; i++) 
            {
                if (n % i == 0) 
                {
                    if (n / i <= k) 
                    {
                        f = 1;
                        ans = i;
                        break;
                    }
                    else 
                    {
                        if (i <= k) 
                        {
                            ans = min(ans, n / i);
                        }
                    }
                }
            }
            cout << ans << endl;
        }
    }
    return 0;
}