CF_Educational Codeforces Round 28_A

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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：给一个长度为n的01序列，现让你从这个序列中删除一些数字，使得1后面没有0，求这个序列最多还剩多少个数字

思路：

一开始题意理解错了，以为只能删除0，于是wa了几发。。。

这道题有两种做法。。。

脑洞和思维：

先存在a[n]里面，遍历a[n]，对于a[i]如果是0则不删除，如果是1则删除，定义最后可以剩余的个数为sum，每次取sum的最大值那么对这两种情况进行讨论

1.a[i]=0，不删，显然sum要加上i和i之前0的个数，然后加上i之后1的个数

2.a[i]=1，删，显然sum要加上i之前0的个数，然后加上i之后1的个数

加之前的0比较好理解，但是为什么只加之后的1呢？这样不会错解吗。

对于第一种情况，假如紧跟着i的是0，那么i循环到i+1的时候必然会有sum_(i+1)>sum_i，即一定可以在后面得到正解，如果紧跟着i的是1那么可以在第二种情况得到正解

对于第二种情况又可以转换到第一种

AC代码

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <cmath>
#include <queue>
#include <algorithm>
#include <vector>
#include <stack>
#define INF 0x3f3f3f3f
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
int n;
int a[105];
int main()
{
    while(~scanf("%d", &n))
    {
        int cnt=0;
        for(int i=0; i<n; i++)
        {
            scanf("%d", &a[i]);
            cnt+=a[i];
        }
        for(int i=0; i<n; i++)
        {
            int sum=0;
            for(int j=0; j<n; j++)
                sum+=(j<=i?!a[j]:a[j]);
            cnt=max(cnt, sum);
        }
        printf("%d\n", cnt);
    }
    return 0;
}

动态规划:

可以设置dp[i][j]是前i个数中以j为结尾所满足的序列的最大长度

那么可以根据第i位的值来进行决策，具体看代码

AC代码

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <cmath>
#include <queue>
#include <algorithm>
#include <vector>
#include <stack>
#define INF 0x3f3f3f3f
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
int n;
int a[105];
int dp[105][2];
int main()
{
    while(~scanf("%d", &n))
    {
        memset(a, -1, sizeof(a));
        for(int i=0; i<n; i++)
            scanf("%d", &a[i]);
        memset(dp, 0, sizeof(dp));
        if(a[0])
            dp[0][1]=1;
        else
            dp[0][0]=1;
        for(int i=1; i<n; i++)
        {
            if(a[i])  //根据a[i]的值进行决策
            {
                dp[i][1]=max(dp[i-1][1], dp[i-1][0])+1;
                dp[i][0]=dp[i-1][0];
            }
            else
            {
                dp[i][1]=dp[i-1][1];
                dp[i][0]=dp[i-1][0]+1;
            }
        }
        printf("%d\n", max(dp[n-1][0], dp[n-1][1]));
    }
}