最长递增子序列子串_最长递增奇偶子序列

最长递增子序列子串

Problem statement:

问题陈述：

Given a sequence of numbers you have to find out the length of the longest increasing odd even subsequence and print the length of the subsequence. The sequence will be maintaining like, (odd ) -> ( even ) -> ( odd ) -> ( even ) or (even ) -> ( odd ) -> ( even ) -> ( odd ) .

给定一个数字序列，您必须找出最长的递增奇数偶数子序列的长度并打印该子序列的长度。顺序将保持为(odd)->(even)->(奇数)->(偶数)或(even)->(奇数)->(偶数)->(奇数) 。

Input:
T Test case
T no. of input array along with their element no. N
E.g.
3
8
2 3 4 8 2 5 6 8
8
2 3 4 8 2 6 5 4
7
6 5 9 2 10 77 5
Constrain:
1≤ T ≤ 20
1≤ N ≤50
1≤ A[i] ≤50
Output:
Print the length of the longest increasing 
odd even subsequence

Example

例

T=3
Input:
8
2 3 4 8 2 5 6 8 
Output:
5 ( 2 3 4 5 8  )
Input:
8
2 3 4 8 2 6 5 4
Output:
4 ( 2 3 4 5 )
Input:
7
6 5 9 2 10 77 5
Output:
4 (6 9 10 77 )

Explanation with example:

举例说明：

Let N be the number of elements say, X₁, X₂, X₃ ... X_n.

令N为元素数，即X ₁ ，X ₂ ，X ₃ ... X _{n 。}

Let odd(a) = the value at the index a of the odd array and even(a) = the value at the index a of the even array.

令odd(a) =奇数数组的索引a处的值，而even(a) =偶数数组的索引a处的值。

To find the length of the longest increasing odd even subsequence we will follow these steps,

要找到最长的递增奇数偶数子序列的长度，我们将按照以下步骤操作，

We take two new array one is an odd array and another is even an array and initialize both with 1. We start our algorithm with the second column. We check elements that are before the current element, with the current element.
我们采用两个新的数组，一个是奇数数组，另一个是偶数数组，并都用1初始化。我们从第二列开始我们的算法。我们使用当前元素检查当前元素之前的元素。
If the current element is odd and the comparing the element is even then,
如果当前元素为奇数，而比较元素为偶数，

odd (index of current element) = even (index of the comparing element) + 1 ;
奇数(当前元素的索引)=偶数(比较元素的索引)+1；
If the current element is even and the comparing element is odd then,
如果当前元素为偶数，而比较元素为奇数，

even (index of current element) = odd (index of the comparing element) + 1;
偶数(当前元素的索引)=奇数(比较元素的索引)+1；

C++ Implementation:

C ++实现：

#include <bits/stdc++.h>
using namespace std;
int length_of_subsequence(int* arr, int n)
{
int a[n];
for (int i = 0; i < n; i++) {
a[i] = 1;
}
for (int i = 1; i < n; i++) {
for (int j = 0; j < i; j++) {
if (arr[i] % 2 == 0) {
if (arr[j] % 2 == 1 && arr[j] < arr[i]) {
a[i] = max(a[i], a[j] + 1);
}
}
else {
if (arr[j] % 2 == 0 && arr[j] < arr[i]) {
a[i] = max(a[i], a[j] + 1);
}
}
}
}
int Max = 0;
for (int i = 0; i < n; i++) {
Max = max(Max, a[i]);
}
cout << endl;
return Max;
}
int main()
{
int t;
cout << "TestCase : ";
cin >> t;
while (t--) {
int n;
cout << "Enter number of elements : ";
cin >> n;
int arr[n];
cout << "Enter the elements : ";
for (int i = 0; i < n; i++) {
cin >> arr[i];
}
cout << "Length of the subsequence : " << length_of_subsequence(arr, n) << endl;
}
return 0;
}

Output

输出量

TestCase : 3
Enter number of elements : 8
Enter the elements : 2 3 4 8 2 5 6 8
Length of the subsequence : 5
Enter number of elements : 8
Enter the elements : 2 3 4 8 2 6 5 4
Length of the subsequence : 4
Enter number of elements : 7
Enter the elements : 6 5 9 2 10 77 5
Length of the subsequence : 4