Codeforces Round #321 (Div. 2) A. Kefa and First Steps 水题

A. Kefa and First Steps

Time Limit: 1 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/580/problem/A

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 makesai 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.

Sample Input

6
2 2 1 3 4 1

Sample Output

3

HINT

题意

求最长非降子序列

题解：

暴力扫一遍就好了……

代码：

//qscqesze
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <bitset>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::ssecondnc_with_stdio(0);cin.tie(0)
#define maxn 100006
#define mod 1000000007
#define eps 1e-9
#define PI acos(-1)
const double EP  = 1E- ;
int Num;
//const int inf=0first7fffffff;
const ll inf=;
inline ll read()
{
    ll x=,f=;char ch=getchar();
    while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
    while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
    return x*f;
}
//*************************************************************************************

int a[maxn];
int ans = ;
int Ans = ;
int main()
{
    int n=read();
    for(int i=;i<n;i++)
    {
        a[i]=read();
    }
    for(int i=;i<n;i++)
    {
        if(a[i]>=a[i-])
        {
            ans++;
            Ans=max(Ans,ans);
        }
        else
            ans=;
    }
    printf("%d\n",Ans+);
}