就是阶梯NIM博弈，那么看层数是不是奇数的异或就行了；#include#include#include#include#include#include#include#include#include#include#include#include#include#include#include#include#include//#include//#pragma GCC optimize(2)using namespace std;#define maxn 400005#define inf 0x7fffffff//#define INF 1e18#define rdint(x) scanf("%d",&x)#define rdllt(x) scanf("%lld",&x)#define rdult(x) scanf("%lu",&x)#define rdlf(x) scanf("%lf",&x)#define rdstr(x) scanf("%s",x)typedef long long ll;typedef unsigned long long ull;typedef unsigned int U;#define ms(x) memset((x),0,sizeof(x))const long long int mod = 1e9;#define Mod 1000000000#define sq(x) (x)*(x)#define eps 1e-11typedef pair pii;#define pi acos(-1.0)//const int N = 1005;#define REP(i,n) for(int i=0;i<(n);i++)typedef pair pii;inline int rd() { int x = 0; char c = getchar(); bool f = false; while (!isdigit(c)) { if (c == '-') f = true; c = getchar(); } while (isdigit(c)) { x = (x << 1) + (x << 3) + (c ^ 48); c = getchar(); } return f ? -x : x;}ll gcd(ll a, ll b) { return b == 0 ? a : gcd(b, a%b);}int sqr(int x) { return x * x; }/*ll ans;ll exgcd(ll a, ll b, ll &x, ll &y) { if (!b) { x = 1; y = 0; return a; } ans = exgcd(b, a%b, x, y); ll t = x; x = y; y = t - a / b * y; return ans;}*/int T;vectorvc[maxn];int ans;int a[maxn];void dfs(int u,int fa,int dep){ for(int i=0;i int v=vc[u][i]; if((dep+1)%2)ans^=a[v]; dfs(v,u,dep+1); }}int main() { T=rd(); while(T--){ int n;rdint(n); for(int i=0;i<=n;i++)vc[i].clear(); for(int i=1;i int fa;rdint(fa); vc[fa].push_back(i); } ans=0; for(int i=0;i dfs(0,0,0); if(ans)cout<<"win"< else cout<<"lose"< }} EPFL - Fighting
