acm寒假特辑 1月18日 CodeForces

C - 3 CodeForces - 580A

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.

题目大意：求递增区间长度。e.g.（1 3 4） 上题答案为3。
思路：先输入，再比较，同时计数x++，与此同时与max比较，实时更新max的大小，遇到不连续的时候则x重置为1；

#include<iostream>
using namespace std;
int main()
{
    int n, a[100005],x = 1, max=0;
    cin >> n;
    for (int i=0;i<n;i++)
    {
        cin >> a[i];
    }
    for (int i = 0; i < n-1; i++)
    {
        if (a[i] <= a[i + 1])
        {
            x++; 
        }
        else
        {
            if (x>max)
            {
                max = x;
            }
            x = 1;
        }
    }//考虑最后
    if (x >= max)
        max = x;
    cout << max;
    return 0;
}