CodeForces - 702A Maximum Increase （dp最长严格递增子串）

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该博客探讨了CodeForces中702A问题，即寻找给定数组的最大长度递增子数组。博主通过动态规划（dp）的方法来解决此问题，其中dp[i]表示以i为结束的最长严格递增子串的长度。文章介绍了如何利用递推公式找到解决方案，并提供了详细思路。

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题目链接：http://codeforces.com/problemset/problem/702/A点击打开链接

You are given array consisting of n integers. Your task is to find the maximum length of an increasing subarray of the given array.

A subarray is the sequence of consecutive elements of the array. Subarray is called increasing if each element of this subarray strictly greater than previous.

Input

The first line contains single positive integer n (1 ≤ n ≤ 105) — the number of integers.

The second line contains n positive integers a1, a2, ..., an (1 ≤ ai ≤ 109).

Output

Print the maximum length of an increasing subarray of the given array.

   Examples
 
     input
   
5
1 7 2 11 15

     output
   
3

     input
   
6
100 100 100 100 100 100

     output
   
1

     input
   
3
1 2 3

     output
   
3

最长严格递增子串跟求最大区间和很像 dp[i]记录以i结尾的上升的最长子串个数

因为连续的所以很好找到递推方程

#include <iostream>
#include <queue>
#include <stdio.h>
#include <stdlib.h>
#include <stack>
#include <limits.h>
#include <string>
#include <string.h>
#include <vector>
#include <set>
#include <map>
#include <algorithm>
#include <math.h>
#define maxn 101000
using namespace std;
int dp[maxn];
int a[maxn];
int main()
{
    int n=0;
    int maxx=1;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        scanf("%d",&a[i]);
    dp[1]=1;
    for(int i=2;i<=n;i++)
        {
            if(a[i]>a[i-1])
                dp[i]=dp[i-1]+1;
            else
                dp[i]=1;
            maxx=max(maxx,dp[i]);
        }
    printf("%d",maxx);
}