CF - 580A. Kefa and First Steps - dp思维

最新推荐文章于 2021-03-03 16:07:35 发布

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本文介绍了一道关于寻找最长连续非递减子序列的问题。通过输入一系列整数，任务是找出该序列中最大非递减子序列的长度。文章提供了一个简洁有效的解法，采用动态规划的思想，记录当前递增的值，最终输出最大连续非递减子序列的长度。

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1.题目描述：

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 sequence ai. 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.

Examples

input
6
2 2 1 3 4 1

output
3

input
3
2 2 9

output
3

Note

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.

2.题意概述：

给你一组数组，要你最长的连续非递减子串的长度。

3.解题思路：

既然是连续，考虑dp思想，以i结尾最大的连续子串长度就和当前a[i]的值相关。因此记录一下当前递增的值即可。

4.AC代码：

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
#define maxn 100100
#define lson root << 1
#define rson root << 1 | 1
#define lent (t[root].r - t[root].l + 1)
#define lenl (t[lson].r - t[lson].l + 1)
#define lenr (t[rson].r - t[rson].l + 1)
#define N 1111
#define eps 1e-6
#define pi acos(-1.0)
#define e exp(1.0)
using namespace std;
const int mod = 1e9 + 7;
typedef long long ll;
typedef unsigned long long ull;
int a[maxn];
int main()
{
#ifndef ONLINE_JUDGE
	freopen("in.txt", "r", stdin);
	freopen("out.txt", "w", stdout);
	long _begin_time = clock();
#endif
	int n;
	while (~scanf("%d", &n))
	{
		int ans = 0;
		int cur = 0, now = 0;
		for (int i = 0; i < n; i++)
		{
			scanf("%d", &a[i]);
			if (a[i] >= now)
			{
				cur++;
				ans = max(ans, cur);
			}
			else
				cur = 1;
			now = a[i];
		}
		printf("%d\n", ans);
	}
#ifndef ONLINE_JUDGE
	long _end_time = clock();
	printf("time = %ld ms.", _end_time - _begin_time);
#endif
	return 0;
}