Longest Ordered Subsequence(DP)

A numeric sequence of ai is ordered if a1 < a2 < ... < aN. Let the subsequence of the given numeric sequence (a1a2, ..., aN) be any sequence (ai1ai2, ..., aiK), where 1 <= i1 < i2 < ... < iK <= N. For example, sequence (1, 7, 3, 5, 9, 4, 8) has ordered subsequences, e. g., (1, 7), (3, 4, 8) and many others. All longest ordered subsequences are of length 4, e. g., (1, 3, 5, 8).

Your program, when given the numeric sequence, must find the length of its longest ordered subsequence.

Input

The first line of input file contains the length of sequence N. The second line contains the elements of sequence - N integers in the range from 0 to 10000 each, separated by spaces. 1 <= N <= 1000

Output

Output file must contain a single integer - the length of the longest ordered subsequence of the given sequence.

Sample Input

7
1 7 3 5 9 4 8

Sample Output

4

 

#include<stdio.h>
#include<iostream>
#include<string.h>
using namespace std;
const int N=1010;
int a[N];
int f[N];

int main(){
	int i,j,k;
	int sum;
	int t,n,m;
	
	while(cin>>n){
		for(int i=1;i<=n;i++)
		{
			cin>>a[i];
		}
		memset(f,0,sizeof(f));//3
		for(int i=1;i<=n;i++)
		{
			f[i]=1;//1
			for(int j=1;j<i;j++)
			{
				if(a[i]>a[j])//2
				{
					f[i]=max(f[i],f[j]+1); 
				}
			}
		}
		int ans=0;
		for(int i=1;i<=n;i++)
		{
			if(ans<f[i])ans=f[i];
		}
		cout<<ans<<endl;
	}
	return 0;
} 

 

在C语言中,Longest Ordered Subsequence (最长递增子序列)是一个常见的动态规划问题。我们可以使用二维数组或者自定义数据结构来解决它。这里是一个简单的解决方案: ```c #include <stdio.h> // 定义一个结构体表示元素和它的索引 typedef struct { int num; int index; } Node; // 动态规划函数,lis[i] 存储以 nums[i] 结尾的最长递增子序列长度 int longestIncreasingSubsequence(int arr[], int n) { int lis[n]; // 初始化所有序列长度为1,因为每个数都是其自身的单元素序列 for (int i = 0; i < n; i++) { lis[i] = 1; } // 遍历数组,比较当前元素与前一个元素 for (int i = 1; i < n; i++) { for (int j = 0; j < i; j++) { if (arr[i] > arr[j]) { // 如果当前元素大于前一个,尝试将其添加到前者的序列中 lis[i] = max(lis[i], lis[j] + 1); } } } // 找到全局最大值即为最长递增子序列的长度 int max_len = 0; for (int i = 0; i < n; i++) { max_len = max(max_len, lis[i]); } return max_len; } // 辅助函数计算两个整数的最大值 int max(int a, int b) { return a > b ? a : b; } int main() { int arr[] = {10, 9, 2, 5, 3, 7, 101, 18}; int n = sizeof(arr) / sizeof(arr[0]); int result = longestIncreasingSubsequence(arr, n); printf("Length of the Longest Increasing Subsequence is %d\n", result); return 0; } ``` 在这个代码中,`longestIncreasingSubsequence`函数通过迭代数组并更新每个元素的最长递增子序列长度,最后返回整个序列中最长的一个。
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