CodeForces 580A Kefa and First Steps

本文介绍了一种求解最长非递减子序列的方法,通过遍历序列并比较相邻元素来确定最大非递减子段长度。适用于解决如Kefa在互联网上连续n天收入的非递减片段问题。

题意:求最长不递减子序列

思路:直接扫一遍就好了...


#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5+6;
int ans = 0;
int a[maxn];
int anss = 0;
int main()
{
     int n;
	 scanf("%d",&n);
	 for (int i = 1;i<=n;i++)
		 scanf("%d",&a[i]);
	 for (int i = 2;i<=n;i++)
	 {
		 if (a[i]>=a[i-1])
		 {
			 anss++;
			 ans = max(ans,anss);
		 }
		 else
			 anss=0;
	 }
	 printf("%d\n",ans+1);
}


Description

Kefa decided to make some money doing business on the Internet for exactly n days. He knows that on the i-th day (1 ≤ i ≤ n) he makes ai money. Kefa loves progress, that's why he wants to know the length of the maximum non-decreasing subsegment in sequenceai. Let us remind you that the subsegment of the sequence is its continuous fragment. A subsegment of numbers is called non-decreasing if all numbers in it follow in the non-decreasing order.

Help Kefa cope with this task!

Input

The first line contains integer n (1 ≤ n ≤ 105).

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

Output

Print a single integer — the length of the maximum non-decreasing subsegment of sequence a.

Sample Input

Input
6
2 2 1 3 4 1
Output
3
Input
3
2 2 9
Output
3

Hint

In the first test the maximum non-decreasing subsegment is the numbers from the third to the fifth one.

In the second test the maximum non-decreasing subsegment is the numbers from the first to the third one.



虽然给定引用中未直接提及“Kuroni and Simple Strings”题目的详细信息,但通常这类题目可能与字符串处理、括号匹配等相关。一般而言,题目可能会给出一个由括号组成的字符串,要求找出能移除的最大数量的不相交的合法括号对,并输出移除这些括号对后的相关信息。 ### 解法分析 #### 栈解法 栈解法是处理括号匹配问题的经典方法。通过遍历字符串,将左括号压入栈中,遇到右括号时,若栈顶为左括号,则将栈顶元素弹出,表示这是一对匹配的括号。 ```python s = input() stack = [] pairs = [] for i, char in enumerate(s): if char == '(': stack.append(i) else: if stack: left_index = stack.pop() pairs.append((left_index + 1, i + 1)) if not pairs: print(0) else: print(1) print(len(pairs) * 2) result = [] for l, r in pairs: result.extend([l, r]) result.sort() print(" ".join(map(str, result))) ``` #### 双指针解法 双指针解法从字符串的两端向中间遍历,分别使用两个指针 `left` 和 `right`。`left` 指针从左向右寻找 `(`,`right` 指针从右向左寻找 `)`,当找到一对匹配的括号时,将它们标记为已移除,继续寻找下一对匹配的括号,直到无法再找到匹配的括号为止。 ```python s = input() n = len(s) left = 0 right = n - 1 pairs = [] while left < right: while left < right and s[left] != '(': left += 1 while left < right and s[right] != ')': right -= 1 if left < right: pairs.append((left + 1, right + 1)) left += 1 right -= 1 if not pairs: print(0) else: print(1) print(len(pairs) * 2) result = [] for l, r in pairs: result.extend([l, r]) result.sort() print(" ".join(map(str, result))) ``` ### 复杂度分析 - **栈解法**:时间复杂度为 $O(n)$,其中 $n$ 是字符串的长度。空间复杂度为 $O(n)$,主要用于栈的空间开销。 - **双指针解法**:时间复杂度为 $O(n)$,空间复杂度为 $O(n)$,主要用于存储匹配的括号对。
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