F - Select Half

Time Limit: 2 sec / Memory Limit: 1024 MB

思路：n按奇数偶数分开写，如果n是偶数，中途最多可以有两个不选，设置dp1,dp2，

　　　如果n是奇数，中间最多有三个可以不选，设置三个数组，算了我不知道怎么用语言解释

Score :$600$ points

Problem Statement

Given is an integer sequence ${\mathrm{A1,...,AN}}^{}$of length $N$.

We will choose exactly $\lfloor N/2\rfloor$elements from this sequence so that no two adjacent elements are chosen.

Find the maximum possible sum of the chosen elements.

Here $\lfloor x\rfloor$denotes the greatest integer not greater than $\mathrm{xx}$.

Constraints

• $\mathrm{2\le N\le 2\times 105}$
• $|Ai|\le 109$
• All values in input are integers.

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Input

Input is given from Standard Input in the following format:

$N$
${\mathrm{A1}}^{}$ $......$ ${\mathrm{AN}}^{}$

Output

Print the maximum possible sum of the chosen elements.

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Sample Input 1 Copy

Copy

6
1 2 3 4 5 6

Sample Output 1 Copy

Copy

12

Choosing $2$, $4$, and $6$makes the sum $12$, which is the maximum possible value.

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Sample Input 2 Copy

Copy

5
-1000 -100 -10 0 10

Sample Output 2 Copy

Copy

0

Choosing $-10$and $10$ makes the sum $0$, which is the maximum possible value.

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Sample Input 3 Copy

Copy

10
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000

Sample Output 3 Copy

Copy

5000000000

Watch out for overflow.

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Sample Input 4 Copy

Copy

27
18 -28 18 28 -45 90 -45 23 -53 60 28 -74 -71 35 -26 -62 49 -77 57 24 -70 -93 69 -99 59 57 -49

Sample Output 4 Copy

Copy

295

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define rep(i,n) for(int i=0;i<n;i++)
#define repr(i,n) for(int i=n-1;i>=0;i--)
const int N=2e5;

int main(){
int n;
cin>>n;
int a[n];
rep(i,n) scanf("%d",&a[i]);
if(n%2==0){
ll dp[n];
dp[0]=a[0];
dp[1]=a[1];
    for(int i=2;i<n;i++){
    if(i%2==0)
        dp[i]=dp[i-2]+a[i];
    else
        dp[i]=max(dp[i-3],dp[i-2])+a[i];
    }
    printf("%lld\n",max(dp[n-1],dp[n-2]));
}
else{
    vector<ll>l(n/2),m(n/2),r(n/2);
    l[0]=a[0];
    m[0]=a[1];
    r[0]=a[2];
    for(int i=1;i<n/2;i++){
        l[i]=l[i-1]+a[2*i];
        m[i]=max(l[i-1],m[i-1])+a[2*i+1];
        r[i]=max(l[i-1],max(m[i-1],r[i-1]))+a[2*i+2];
    }
    printf("%lld\n",max(l[n/2-1],max(m[n/2-1],r[n/2-1])));
}

return 0;
}