題目描述
There are n models in the shop numbered from 1 to n, with sizes s1,s2,…,sn.
Orac will buy some of the models and will arrange them in the order of increasing numbers (i.e. indices, but not sizes).
Orac thinks that the obtained arrangement is beatiful, if for any two adjacent models with indices ij and ij+1 (note that ij<ij+1, because Orac arranged them properly), ij+1 is divisible by ij and sij<sij+1.
For example, for 6 models with sizes {3,6,7,7,7,7}, he can buy models with indices 1, 2, and 6, and the obtained arrangement will be beautiful. Also, note that the arrangement with exactly one model is also considered beautiful.
Orac wants to know the maximum number of models that he can buy, and he may ask you these queries many times.
Input
The first line contains one integer t (1≤t≤100): the number of queries.
Each query contains two lines. The first line contains one integer n (1≤n≤100000): the number of models in the shop, and the second line contains n integers s1,…,sn (1≤si≤109): the sizes of models.
It is guaranteed that the total sum of n is at most 100000.
Output
Print t lines, the i-th of them should contain the maximum number of models that Orac can buy for the i-th query.
Example
input
4
4
5 3 4 6
7
1 4 2 3 6 4 9
5
5 4 3 2 1
1
9
output
2
3
1
1
Note
In the first query, for example, Orac can buy models with indices 2 and 4, the arrangement will be beautiful because 4 is divisible by 2 and 6 is more than 3. By enumerating, we can easily find that there are no beautiful arrangements with more than two models.
In the second query, Orac can buy models with indices 1, 3, and 6. By enumerating, we can easily find that there are no beautiful arrangements with more than three models.
In the third query, there are no beautiful arrangements with more than one model.
題目大意
給出n個數，找到滿足條件的子序列的最大長度。
條件：1.保證a[ i[j] ]<a[ i[j+1] ]。2.i[j]可以整除i[j+1]。
題目分析
枚舉所有的起點進行dfs。dfs的深度即爲得到答案的長度，找出最大值即可。
注意：因爲上一層找到的數組的下標要能整除這一層找到數組的下標。因此循環時可以只找上一次下標u的倍數。
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 <unordered_set>
#include <algorithm>
#include <iomanip>
#define LL long long
using namespace std;
int const N=1e5+5;
int n,a[N];
int len=0; //長度
int dfs(int u,int cnt) //dfs求解答案（u爲上一層找到的數的下標，cnt爲深度）
{
len=max(len,cnt); //記錄最大值
for(int i=u;i<=n;i+=u) //枚舉u的倍數找下一層
{
if(a[u]<a[i]) dfs(i,cnt+1); //滿足條件則繼續進入下一層
}
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
for(int i=1;i<=n;i++)
scanf("%d",&a[i]);
len=0;
for(int i=1;i<=n;i++) //枚舉所有起點
dfs(i,1);
cout<<len<<endl; //輸出答案（長度）
}
return 0;
}