[Codeforces Round #659 (Div. 1)] B. GameGame （博弈论）

[Codeforces Round #659 (Div. 1)] B. GameGame （博弈论）

思路：

\(cnt_i\)代表这\(\mathit n\)个数中有多少个数在二进制表示中第\(\mathit i\)位为\(\text 1\)。

我们知道如果\(cnt_i \%2=0\)，无论如何取，先手和后手在第\(\mathit i\)位的值都相等。

所以要找到最大的\(id\)，使其\(cnt_{id} \%2=1\)。先手和后手的最终分数在二进制表示法中第\(id\)位一定不同。

所以玩家的胜败只在第\(id\)位（因为第\(id\)位的权值大，后面的无论咋取不影响两者大小关系。）决定。

当\((cnt_{id}-1)/2\)是偶数，那么先手必赢。

决策为：先手在该位先取一个\(\text 1\)，然后跟着后手选，那么后手会选择偶数个\(\text 1\)，结果为：先手在该位值为\(\text 1\)，后手为\(\text 0\)。

当\((cnt_{id}-1)/2\)是奇数：

​ 如果\(n\%2=1\)，先手必输：

​ 决策为：每一个回合后手只需要跟着先手选一样的，那么先手会选择偶数次\(\text 1\)，结果为\(\text 0\)，后手反之。

​ 如果\(n\%2=0\)，先手必赢：

​ 决策为：先手第一步选一个\(\text 0\)，然后将状态转为\(n\%2=1\)的局面，且轮到后手先选择，上面已经证明，这种局面先选择的必输。

代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <bits/stdc++.h>
#define ALL(x) (x).begin(), (x).end()
#define sz(a) int(a.size())
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define chu(x)  if(DEBUG_Switch) cout<<"["<<#x<<" "<<(x)<<"]"<<endl
#define du3(a,b,c) scanf("%d %d %d",&(a),&(b),&(c))
#define du2(a,b) scanf("%d %d",&(a),&(b))
#define du1(a) scanf("%d",&(a));
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
ll powmod(ll a, ll b, ll MOD) { if (a == 0ll) {return 0ll;} a %= MOD; ll ans = 1; while (b) {if (b & 1) {ans = ans * a % MOD;} a = a * a % MOD; b >>= 1;} return ans;}
ll poww(ll a, ll b) { if (a == 0ll) {return 0ll;} ll ans = 1; while (b) {if (b & 1) {ans = ans * a ;} a = a * a ; b >>= 1;} return ans;}
void Pv(const vector<int> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%d", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("\n");}}}
void Pvl(const vector<ll> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%lld", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("\n");}}}
inline long long readll() {long long tmp = 0, fh = 1; char c = getchar(); while (c < '0' || c > '9') {if (c == '-') fh = -1; c = getchar();} while (c >= '0' && c <= '9') tmp = tmp * 10 + c - 48, c = getchar(); return tmp * fh;}
inline int readint() {int tmp = 0, fh = 1; char c = getchar(); while (c < '0' || c > '9') {if (c == '-') fh = -1; c = getchar();} while (c >= '0' && c <= '9') tmp = tmp * 10 + c - 48, c = getchar(); return tmp * fh;}
void pvarr_int(int *arr, int n, int strat = 1) {if (strat == 0) {n--;} repd(i, strat, n) {printf("%d%c", arr[i], i == n ? '\n' : ' ');}}
void pvarr_LL(ll *arr, int n, int strat = 1) {if (strat == 0) {n--;} repd(i, strat, n) {printf("%lld%c", arr[i], i == n ? '\n' : ' ');}}
const int maxn = 1000010;
const int inf = 0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
#define DEBUG_Switch 0
int n;
int a[maxn];
int cnt[40];
void PRINF2(ll num, int k)
{
    for (int i = k; i >= 0; i--)
    {
        cout << (bool)(num & (1ll << i));
    }
    cout << endl;
}
int main()
{
#if DEBUG_Switch
    freopen("C:\\code\\input.txt", "r", stdin);
#endif
    //freopen("C:\\code\\output.txt","w",stdout);
    int t;
    t = readint();
    while (t--)
    {
        n = readint();
        repd(i, 1, n)
        {
            a[i] = readint();
        }
        // repd(i, 1, n)
        // {
        //     PRINF2(a[i], 10);
        // }
        if (n == 1)
        {
            if (a[1] == 0)
            {
                printf("DRAW\n");
            } else
            {
                printf("WIN\n");
            }
        } else if (n == 2)
        {
            if (a[1] == a[2])
            {
                printf("DRAW\n");
            } else
            {
                printf("WIN\n");
            }
        } else
        {
            int id = -1;
            for (int i = 0; i <= 30; ++i)
            {
                repd(j, 1, n)
                {
                    if (a[j] & (1 << i))
                    {
                        cnt[i]++;
                    }
                }
            }
            for (int i = 30; i >= 0; --i)
            {
                if (cnt[i] % 2 == 1)
                {
                    id = i;
                    break;
                }
            }
            if (id == -1)
            {
                printf("DRAW\n");
            } else
            {
                if ( ((cnt[id] + 1) / 2) % 2 == 1)
                {
                    printf("WIN\n");
                } else
                {
                    if (n & 1)
                        printf("LOSE\n");
                    else
                        printf("WIN\n");
                }
            }
        }
        for (int i = 30; i >= 0; --i)
            cnt[i] = 0;
    }
    return 0;
}