AcWing 蛋糕游戏

6118. 蛋糕游戏 - AcWing题库

a先手，每回合合并两个蛋糕,当只剩下一个蛋糕时最后吃掉

b后手,每回合吃最左边或最右边的蛋糕

求ab能吃的最多的值

a最多能吃n/2+1个,b吃n/2-1个,如果吃到a合并的蛋糕，获得的值还能更大，因此让a吃最小的n/2+1个，b获得剩下的

#include <bits/stdc++.h>        

#define fi first
#define se second
#define endl '\n'

using namespace std;
using LL = long long;


const int mod = 1e9 + 7;


const int N = 5e5 + 10,T = 20;


int n;
LL a[N];

void solve()
{
    cin >> n;
    int len = n / 2 + 1;

    for (int i = 1;i <= n;i ++) cin >> a[i],a[i] += a[i - 1];

    LL sa = a[n];

    for (int i = 1;i + len - 1<= n;i ++)
    {
        int j = i + len - 1;
        sa = min(sa,a[j] - a[i - 1]);   
    }

    cout << sa << " " << a[n] - sa << endl;

}

int main()
{
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int _ = 1;
    cin >> _;
    while(_--) solve();

    return 0;
}