codeforces 846A Curriculum Vitae

最新推荐文章于 2021-08-03 18:05:50 发布

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最新推荐文章于 2021-08-03 18:05:50 发布

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分类专栏： codeforces 文章标签：模拟

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本文介绍了一个算法问题，旨在帮助游戏制作人小岛秀夫通过删除部分游戏记录来优化他的简历，确保没有不成功的项目紧随成功项目之后，同时尽可能多地保留项目。

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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
4
1 1 0 1

output
3

input
6
0 1 0 0 1 0

output
4

input
1
0

output
1

题意：给n个只含0 1的数列，可以删去任意位置的数，使的数列满足0在1的左边（即没有1在0左边），求满足条件的序列的最大长度，题目看了半天才看懂

模拟一下计算满足条件的最大值即可

#pragma comment(linker,"/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<string>
#include<stack>
#include<queue>
#include<deque>
#include<set>
#include<map>
#include<cmath>
#include<vector>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;

#define pi acos(-1.0)
#define eps 1e-10
#define pf printf
#define sf scanf
#define lson rt<<1,l,m
#define rson rt<<1|1,m+1,r
#define e tree[rt]
#define _s second
#define _f first
#define all(x) (x).begin,(x).end
#define mem(i,a) memset(i,a,sizeof i)
#define for0(i,a) for(int (i)=0;(i)<(a);(i)++)
#define for1(i,a) for(int (i)=1;(i)<=(a);(i)++)
#define mi ((l+r)>>1)
#define sqr(x) ((x)*(x))

const int inf=0x3f3f3f3f;
int n,a[110],ans;

int main()
{
    while(~sf("%d",&n))
    {
        ans=0;
        for1(i,n)sf("%d",&a[i]);
        for1(i,n+1)//在最后加上一个0再计算不会影响结果，防止全是0
        {
            int cnt=0;
            for1(j,n)
            {
                if(j<i&&!a[j])cnt++;//小于第i位是0的情况
                if(j>=i&&a[j])cnt++;//大于等于i位是1的情况
            }
            ans=max(ans,cnt);
        }
        pf("%d\n",ans);
    }
    return 0;
}