题意：给定一个数组，要求Alice和Bob两个人一前一后的选择数并删除，Alice能否选出偶数总和的数以获得胜利。

思路：Alice选偶数时，Bob总是也选偶数留下奇数使得Alice的数字改变奇偶性，Alice选奇数时，Bob总是也选奇数使得Alice选偶数不改变她的奇偶性。

于是我们把奇偶分开来看，对于奇数的数量，做分类讨论

cnt1%4==0,Alice胜

cnt1%4==1，此时cnt2&1，Bob胜，因为剩下一个奇数Alice选，反之Alice胜

cnt1%4==2，此时Bob必胜，因为最后两个奇数一人一个

cnt1%4==3，Alice必胜，因为Alice最后拿两个奇数

#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
#include<map>
#include<set>
#include<vector>
#include<algorithm>
#include<string>
#include<bitset>
#include<cmath>
#include<array>
#include<atomic>
#include<sstream>
#include<stack>
#include<iomanip>
//#include<bits/stdc++.h>//#define int ll
#define IOS std::ios::sync_with_stdio(false);std::cin.tie(0);
#define pb push_back
#define endl '\n'
#define x first
#define y second
#define Endl endl
#define pre(i,a,b) for(int i=a;i<=b;i++)
#define rep(i,b,a) for(int i=b;i>=a;i--)
#define si(x) scanf("%d", &x);
#define sl(x) scanf("%lld", &x);
#define ss(x) scanf("%s", x);
#define YES {puts("YES");return;}
#define NO {puts("NO"); return;}
#define all(x) x.begin(),x.end()using namespace std;typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;
typedef pair<int, PII> PIII;
typedef pair<char, int> PCI;
typedef pair<int, char> PIC;
typedef pair<double, double> PDD;
typedef pair<ll, ll> PLL;
const int N = 1000010, M = 2 * N, B = N, MOD = 998244353;
const double eps = 1e-7;
const int INF = 0x3f3f3f3f;
const ll LLINF = 0x3f3f3f3f3f3f3f3f;//int dx[4] = { -1,0,1,0 }, dy[4] = { 0,1,0,-1 };
int dx[8] = { 1,2,2,1,-1,-2,-2,-1 }, dy[8] = { 2,1,-1,-2,-2,-1,1,2 };
int n;
int a[N];
bool dp[110][110];ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
ll lowbit(ll x) { return x & -x; }
ll qmi(ll a, ll b, ll MOD) {ll res = 1;while (b) {if (b & 1) res = res * a % MOD;a = a * a % MOD;b >>= 1;}return res;
}inline void init() {}void slove()
{cin >> n;pre(i, 1, n) cin >> a[i];int cnt1 = 0, cnt2 = 0;pre(i, 1, n){if (a[i] & 1) cnt1++;else cnt2++;}if (cnt1 % 4 == 0)puts("Alice");else if (cnt1 % 4 == 1){if (n & 1)puts("Bob");else puts("Alice");}else if (cnt1 % 4 == 2)puts("Bob");else if (cnt1 % 4 == 3)puts("Alice");return;
}signed main()
{//IOS;int _ = 1;si(_);init();while (_--){slove();}return 0;
}
/*
1
3 15
1 9
1 9
1 8*/

dp做法

显然，该题的胜负只与奇数和偶数的数量有关，因此，我们做关于奇数和偶数的dp

用odd，even两个二维数组保存当奇数偶数分别为i,j时，先手能否一定取到奇数或者偶数。

哎，这时候就有人要问了，不是只要求能否取到偶数吗，为什么还要有奇数呢？

我们来看看推导过程，首先，当我们从ij向之前的状态转移时，于是我们要的答案变成cntodd&1==1时，even[i][j-1],even[i-1][j]均为false，cntodd&1==0时odd[i-1][j],odd[i][j-1]均为false，所以需要odd辅助，同时，我们也需要维护odd数组

还有需要注意的边界问题（i==0||j==0）

#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
#include<map>
#include<set>
#include<vector>
#include<algorithm>
#include<string>
#include<bitset>
#include<cmath>
#include<array>
#include<atomic>
#include<sstream>
#include<stack>
#include<iomanip>
//#include<bits/stdc++.h>//#define int ll
#define IOS std::ios::sync_with_stdio(false);std::cin.tie(0);
#define pb push_back
#define endl '\n'
#define x first
#define y second
#define Endl endl
#define pre(i,a,b) for(int i=a;i<=b;i++)
#define rep(i,b,a) for(int i=b;i>=a;i--)
#define si(x) scanf("%d", &x);
#define sl(x) scanf("%lld", &x);
#define ss(x) scanf("%s", x);
#define YES {puts("YES");return;}
#define NO {puts("NO"); return;}
#define all(x) x.begin(),x.end()using namespace std;typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;
typedef pair<int, PII> PIII;
typedef pair<char, int> PCI;
typedef pair<int, char> PIC;
typedef pair<double, double> PDD;
typedef pair<ll, ll> PLL;
const int N = 1000010, M = 2 * N, B = N, MOD = 998244353;
const double eps = 1e-7;
const int INF = 0x3f3f3f3f;
const ll LLINF = 0x3f3f3f3f3f3f3f3f;//int dx[4] = { -1,0,1,0 }, dy[4] = { 0,1,0,-1 };
int dx[8] = { 1,2,2,1,-1,-2,-2,-1 }, dy[8] = { 2,1,-1,-2,-2,-1,1,2 };
int n;
int a[N];
bool even[110][110];
bool odd[110][110];ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
ll lowbit(ll x) { return x & -x; }
ll qmi(ll a, ll b, ll MOD) {ll res = 1;while (b) {if (b & 1) res = res * a % MOD;a = a * a % MOD;b >>= 1;}return res;
}inline void init() {/*for (int i = 1; i <= 100; i++){even[i][0] = (i + 1) / 2 & 1 ^ 1;even[0][i] = 1;odd[i][0] = (i + 1) / 2 & 1;odd[0][i] = 0;}*/even[0][1] = true, even[1][0] = false;odd[1][0] = true, odd[0][1] = false;for(int cnt=2;cnt<=100;cnt++)for (int i = 0; i <= cnt; i++) {int j = cnt - i;if (i & 1){if(i) even[i][j] |= !even[i - 1][j];if(j) even[i][j] |= !even[i][j - 1];if(i) odd[i][j] |= !odd[i - 1][j];if(j) odd[i][j] |= !odd[i][j - 1];}else{if(i) even[i][j] |= !odd[i - 1][j];if(j) even[i][j] |= !odd[i][j - 1];if(i) odd[i][j] |= !even[i - 1][j];if(j) odd[i][j] |= !even[i][j - 1];}}
}void slove()
{cin >> n;pre(i, 1, n) cin >> a[i];int cnt1 = 0, cnt2 = 0;pre(i, 1, n){if (a[i] & 1) cnt1++;else cnt2++;}if (even[cnt1][cnt2])puts("Alice");else puts("Bob");
}signed main()
{//IOS;int _ = 1;si(_);init();while (_--){slove();}return 0;
}
/*
1
3 15
1 9
1 9
1 8*/

这是1500的博弈题？，这就叫惊喜！

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dp做法

这是1500的博弈题？，这就叫惊喜！

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