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

1.题目描述：

A. Kefa and First Steps time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output

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; }