题意：
给你一个小根堆，从根开始拿，拿走子节点被拿完后才可以拿走父节点，两个人依次拿，谁拿的节点总和大谁获胜，问你谁有必胜策略。
题解：
小根堆中，每个点的权值总是不小于父亲节点的权值。所以无论怎么取，先拿走的数一定 不小于后面拿走的数。 此时双方的最优策略就是：贪心选择能取的数字之中最大的数。
#include<iostream> #include<cstdio> #include<cstring> #include<cmath> #include<string> #include<stack> #include<algorithm> using namespace std; #define INF 0x3f3f3f3f #define MAXN 100000+50 #define MAXM 30000 #define ll long long #define rep(i,n,m) for(int i=n;i<=m;++i) #define mod 1000000000 + 7 #define mian main #define mem(a, b) memset(a, b, sizeof a) #ifndef ONLINE_JUDGE #define dbg(x) cout << #x << "=" << x << endl; #else #define dbg(x) #endif inline int read() { int x = 0, f = 1; char ch = getchar(); while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getchar(); } while (ch >= '0' && ch <= '9') { x = 10 * x + ch - '0'; ch = getchar(); } return x * f; } inline ll readll() { ll x = 0, f = 1; char ch = getchar(); while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getchar(); } while (ch >= '0' && ch <= '9') { x = 10 * x + ch - '0'; ch = getchar(); } return x * f; } ll a[MAXN]; int main() { int T = read(); while (T--) { int n = read(); ll sums = 0, sume = 0; rep(i, 1, n) a[i] = readll(); sort(a + 1, a + n + 1); for (int i = n; i >= 1; i-=2) { sums += a[i]; } for (int i = n - 1; i >= 1; i -= 2) { sume += a[i]; } printf("%lld %lld ", sums, sume); } return 0; }