A. Submission Bait

 思路:

如果最大值有奇数个显然Alice赢，否则只需要看排序后是否存在n−i+1是否为奇数且ai>ai−1即可。

 代码:

#include<bits/stdc++.h>
using namespace std;
#define N 2000010
typedef long long ll;
typedef unsigned long long ull;
ll n, m, h, k, t, x, y, z;
ll a, b, c, d, mod = 998244353;
ll ans, num, sum1=0, sum, sum2=0, cnt, maxx, minn = 1e9;
ll f1[N], f2[N], dp[N], an[N];
bool flag, vis[N];
string s,s1,s2;
map<ll, ll>mp;
bool cmp(ll x, ll y) {
	return x > y;
}
//void solve() {
//	cin >> n >> x >> y;
//
//	if (x >= y) {
//		for (int i = 1; i <= n; i++) {
//			if(dp[i]>=x&&)
//		}
//	}
//	else {
//		for (int i = x + 1; i < y; i++)
//			dp[i] = -1;
//	}
//	for (int i = 1; i <= n; i++)
//		cout << dp[i] << " ";
//	cout << endl;
//}
void solve() {
    int n;
    cin >> n;
    vector<int>a(n);
    vector<int>vis(55);
    for (int i = 0; i < n; i++) cin >> a[i], vis[a[i]]++;
    sort(a.begin(), a.end());
    int mx = a[n - 1];
    int ans = 0;
    for (int i = n - 1; i >= 0; i--)
        if (a[i] == mx) ans++;
    if (ans % 2 == 1)
    {
        cout << "YES\n";
        return;
    }
    for (int i = n; i >= 1; i--)
    {
        if (vis[i] % 2 == 1) {
            cout << "YES\n";
            return;
        }
    }

    cout << "NO\n";
}
int main()
{
	cin >> t;
	while (t--) {
		solve();
	}
}

B. Array Craft

 思路:

可以确定x和y这两个位置一定是1且x+1和y-1一定是-1，不难想到可以把y到x之间全填1，两边1和-1交叉填即可.

代码:

#include<bits/stdc++.h>
using namespace std;
#define N 2000010
typedef long long ll;
typedef unsigned long long ull;
ll n, m, h, k, t, x, y, z;
ll a, b, c, d, mod = 998244353;
ll ans, num, sum1=0, sum, sum2=0, cnt, maxx, minn = 1e9;
ll f1[N], f2[N], dp[N], an[N];
bool flag, vis[N];
string s,s1,s2;
map<ll, ll>mp;
bool cmp(ll x, ll y) {
	return x > y;
}
void solve() {
	cin >> n >> x >> y;
	for (int i = 1; i <= n; i++) {
		if (i >= y && i <= x) {
			dp[i] = 1;
		}
	}
	for (int i = y-1; i >=1;i-=2) {
		dp[i] = -1, dp[i - 1] = 1;
	}
	for (int i = x + 1; i <= n; i += 2) {
		dp[i] = -1, dp[i + 1] = 1;
	}
	for (int i = 1; i <= n; i++)
		cout << dp[i] << " ";
	cout << endl;
}
int main()
{
	cin >> t;
	while (t--) {
		solve();
	}
}

 

 

C. Mad MAD Sum

 

 思路:

通过模拟可知:第一次操作后,数组中出若干个数字块，且呈升序,其中注意到第一次操作后会有个别的长度为1的数字块.

我们发现从第二次操作开始，每个长度 >= 2的数字块都会向右滑动一位,而第一次操作出现的长度为1的数字块都会被覆盖掉

也就是说，两次操作后就是整个数组向右不断移动的过程,模拟完前两轮依次删除最后一个数，统计贡献即可

代码:

#include<bits/stdc++.h>
#include <unordered_map>
using namespace std;
#define N 200010
typedef long long ll;
typedef unsigned long long ull;
ll n, m, h, k, t, x, y, z;
ll a, b, c, d, mod = 998244353;
ll ans, num, sum1=0, sum, sum2=0, maxx, minn = 1e9;
ll f1[N], f2[N], dp[N], an[N],cnt[N];
bool flag, vis[N];
string s,s1,s2;
map<ll, ll>mp;
vector<ll>aa(N);
bool cmp(ll x, ll y) {
	return x > y;
}
void solve() {
	cin >> n;
    for (int i = 1; i <= n; i++) {
        cin >> dp[i];
    }
    sum = 0;
	for (int i = 1; i <= 2; i++) {
		maxx = 0;
		memset(vis, 0, sizeof(vis));
		for (int j = 1; j <= n; j++) {
			if (vis[dp[j]])
				maxx = max(maxx, dp[j]);
			vis[dp[j]] = true;
			sum += dp[j];
			dp[j] = maxx;
		}
	}
	for (int i = 1; i <= n; i++) {
		sum += (n - i + 1) * dp[i];
	}
    cout << sum << endl;
}
int main()
{
	cin >> t;
	while (t--) {
		solve();
	}
}