Educational Codeforces Round 90 (Rated for Div. 2) D - Maximum Sum on Even Positions dp

#include<map>
#include<queue>
#include<time.h>
#include<limits.h>
#include<cmath>
#include<ostream>
#include<iterator>
#include<set>
#include<stack>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define rep(i,m,n) for(int i=m;i<=n;i++)
#define mem(st) memset(st,0,sizeof st)
int read()
{
    int res=0,ch,flag=0;
    if((ch=getchar())=='-')             //判断正负
        flag=1;
    else if(ch>='0'&&ch<='9')           //得到完整的数
        res=ch-'0';
    while((ch=getchar())>='0'&&ch<='9')
        res=res*10+ch-'0';
    return flag?-res:res;
}
typedef long long ll;
typedef pair<int,int> pii;
typedef unsigned long long ull;
typedef pair<double,double> pdd;
const int inf = 0x3f3f3f3f;
const int maxn=1e6+10;
ll a[maxn],dp[maxn],s[maxn],x[maxn],y[maxn],n,dp2[maxn];
void solve()
{
    ll n=read();
    ll tmp=0,ans=0;
    for(int i=1; i<=n; i++)
        a[i]=read();
    for(int i=1; i<=n; i++) 
        if(i&1) 
            tmp+=a[i];
    //找一段区间 使得 区间内偶数位的和 减去 奇数位的和 的差值最大 
    for(int i=2; i<=n; i+=2)
        x[i]=a[i]-a[i-1],y[i]=a[i]-a[i+1];
    for(int i=2; i<=n; i+=2) 
        dp[i]=max(0ll,dp[i-2]+x[i]),dp2[i]=max(0ll,dp2[i-2]+y[i]);
    for(int i=2; i<=n; i+=2) 
        ans=max(ans,dp[i]);
    for(int i=2; i<=n-1; i+=2) 
        ans=max(ans,dp2[i]);
    printf("%lld\n",ans+tmp);
    for(int i=1;i<=n;i++) 
        dp[i]=0,dp2[i]=0;
}
signed main()
{
    int t;
    cin>>t;
    while(t--)
        solve();
    return 0;
}