題目描述
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
2
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.
題目大意
你現在要買n個鐵鍬，商店中有k中不同的賣法，依次每一次賣1到k個鐵鍬，現在你只能選擇其中的一種買法，問最少買幾次同一種的買法，使得剛好買到n個鐵鍬。
題目分析
這道題其實就是求約數。
因爲有1-k種買法，每種買法買1-k把鐵鍬，而且只能選一種買法。因此，我們就要找到小於等於k的n的最大約數，所以我們可以先求出n的所有約數，再從大到小遍歷所有的約數，找到小於等於k的最大約數即可。
答案即爲:n/a[k]。
代碼如下
#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <stack>
#include <map>
#include <unordered_map>
#include <queue>
#include <vector>
#include <set>
#include <algorithm>
#include <iomanip>
#define LL long long
using namespace std;
int const N=1e6+5;
int a[N],cnt=0;
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int n,k;
scanf("%d%d",&n,&k);
if(k>=n) {puts("1"); continue;} //k>=n，可直接輸出n
cnt=0;
for(int i=1;i<=n/i;i++) //求n的約數
if(n%i==0)
{
a[cnt++]=i;
if(i!=n/i) a[cnt++]=n/i;
}
sort(a,a+cnt); //將約數進行排序
int u;
for(int i=cnt-1;i>=0;i--) //找到小於等於k的最大約數
if(k>=a[i]) {u=i; break;}
printf("%d\n",n/a[u]);
}
return 0;
}