Rise longest sequence Ⅰ

Rise longest sequence Ⅰ

Rise longest sequence Ⅱ

给定一个长度为N的数列，求数值严格单调递增的子序列的长度最长是多少。

输入格式
第一行包含整数N。

第二行包含N个整数，表示完整序列。

输出格式
输出一个整数，表示最大长度。

数据范围
1≤N≤1000，
−109≤数列中的数≤109
输入样例：
7
3 1 2 1 8 5 6
输出样例：
4

Solution as follows:
farray is used to store the first iend increases the maximum number of the sub-sequence number
f[i]represents ithe number of the maximum number of sub-sequences at the end of rise
recursion formulas readily available:

f[i]=max(f[i],f[j]+1);

Which f[j]represents f[i]a number before, if f[j]<=f[i]it has found a number can rise up to join this sequence being taken f[i]with f[j]+1the maximum value (so a new found +1) on it.

C ++ code:

#include<bits/stdc++.h>
using namespace std;
int a[1001],f[1001];
int main()
{
	int n;
	cin>>n;
	for(int i=1;i<=n;i++)
		cin>>a[i];
	for(int i=1;i<=n;i++)
	{
		f[i]=1;
		for(int j=1;j<=i;j++)
			if(a[j]<a[i])
				f[i]=max(f[i],f[j]+1);
	}
	int ans=-1;
	for(int i=1;i<=n;i++)
		ans=max(ans,f[i]);
	cout<<ans;
	return 0;	 
}

PS: Write problem solution is not easy, point a praise chant walk QwQ ~