Educational Codeforces Round 28 A. Curriculum Vitae（读懂题意后的暴力水题）

A. Curriculum Vitae

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Hideo Kojima has just quit his job at Konami. Now he is going to find a new place to work. Despite being such a well-known person, he still needs a CV to apply for a job.

During all his career Hideo has produced n games. Some of them were successful, some were not. Hideo wants to remove several of them (possibly zero) from his CV to make a better impression on employers. As a result there should be no unsuccessful game which comes right after successful one in his CV.

More formally, you are given an array s1, s2, ..., sn of zeros and ones. Zero corresponds to an unsuccessful game, one — to a successful one. Games are given in order they were produced, and Hideo can't swap these values. He should remove some elements from this array in such a way that no zero comes right after one.

Besides that, Hideo still wants to mention as much games in his CV as possible. Help this genius of a man determine the maximum number of games he can leave in his CV.

Input

The first line contains one integer number n (1 ≤ n ≤ 100).

The second line contains n space-separated integer numbers s1, s2, ..., sn (0 ≤ si ≤ 1). 0 corresponds to an unsuccessful game, 1 — to a successful one.

Output

Print one integer — the maximum number of games Hideo can leave in his CV so that no unsuccessful game comes after a successful one.

Examples

input

41 1 0 1

output

3

input

60 1 0 0 1 0

output

4

input

10

output

1

题解：

昨天因为看国足然后就没打比赛（赢了hhhhhh，武磊牛逼），然后今天上午做这题的时候卡了一会。。主要是题意很难懂

题意：

王大锤有一堆作品，1表示成功，0表示失败，让你从一堆作品里面剔除几个，使得成功作的后面不能有失败作

坑点：

可以把1的作品也剔除掉。。。因为数据很小直接暴力解决了

代码：

#include<iostream>#include<cstring>#include<stdio.h>#include<math.h>#include<string>#include<stdio.h>#include<queue>#include<stack>#include<map>#include<vector>#include<deque>#include<algorithm>using namespace std;#define INF 100861111#define ll long long#define eps 1e-7#define lson k*2#define rson k*2+1int a[105];int main(){    int i,j,n,ans,temp;    scanf("%d",&n);    ans=1;    for(i=0;i<n;i++)        scanf("%d",&a[i]);    for(i=0;i<n;i++)    {        temp=0;        for(j=0;j<=i;j++)        {            if(!a[j])                temp++;        }        for(j=i;j<n;j++)        {            if(a[j])                temp++;        }        ans=max(ans,temp);    }    printf("%d\n",ans);    return 0;}