hdu多校第六场1012 （hdu6645） Stay Real 假博弈，真贪心

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\n", sums, sume);
    }
    return 0;
}