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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;
}