【HDOJ6655】Just Repeat（贪心）

题意：A和B两个人玩游戏，分别有n和m张牌，A的第i张牌是a[i]，B是b[i]

两人轮流出牌，如果一种编号的牌被其中一个人出了另一个人就不能出自己手中这个编号的牌

两人都按最优策略行动，问获胜者

思路：

考场上好像卡常，补题的时候似乎没有

 #include<bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 typedef unsigned int uint;
 typedef unsigned long long ull;
 typedef pair<int,int> PII;
 typedef pair<ll,ll> Pll;
 typedef vector<int> VI;
 typedef vector<PII> VII;
 //typedef pair<ll,ll>P;
 #define N  200010
 #define M  200010
 #define fi first
 #define se second
 #define MP make_pair
 #define pb push_back
 #define pi acos(-1)
 #define mem(a,b) memset(a,b,sizeof(a))
 #define rep(i,a,b) for(int i=(int)a;i<=(int)b;i++)
 #define per(i,a,b) for(int i=(int)a;i>=(int)b;i--)
 #define lowbit(x) x&(-x)
 #define Rand (rand()*(1<<16)+rand())
 #define id(x) ((x)<=B?(x):m-n/(x)+1)
 #define ls p<<1
 #define rs p<<1|1

 const int MOD=1e9+,inv2=(MOD+)/;
       double eps=1e-;
       int INF=1e9;
       int inf=0x7fffffff;
       int dx[]={-,,,};
       int dy[]={,,-,};

 struct node
 {
     int x,y;
 }q[N];

 int a[N],b[N],c[N],s[],d[N][];
 ull k1,k2;

 int read()
 {
    int v=,f=;
    char c=getchar();
    while(c<||<c) {if(c=='-') f=-; c=getchar();}
    while(<=c&&c<=) v=(v<<)+v+v+c-,c=getchar();
    return v*f;
 }

 unsigned long long rng()
 {
     unsigned long long k3 = k1, k4 = k2;
     k1 = k4;
     k3 ^= k3 << ;
     k2 = k3 ^ k4 ^ (k3 >> ) ^ (k4 >> );
     return k2 + k4;
 }

 bool cmp(node a,node b)
 {
     return a.x+a.y>b.x+b.y;
 }

 void solve()
 {
     int n1=read(),n2=read(),p=read();
     if(p==)
     {
         rep(i,,n1) a[i]=read();
         rep(i,,n2) b[i]=read();
     }
      else
      {
          int mod;
          scanf("%llu%llu%d",&k1,&k2,&mod);
          rep(i,,n1) a[i]=rng()%mod;
          scanf("%llu%llu%d",&k1,&k2,&mod);
          rep(i,,n2) b[i]=rng()%mod;
      }
     int t=;
     rep(i,,n1) c[++t]=a[i];
     rep(i,,n2) c[++t]=b[i];
     sort(c+,c+t+);
     t=unique(c+,c+t+)-c-;
     rep(i,,n1)
     {
         a[i]=lower_bound(c+,c+t+,a[i])-c;
         d[a[i]][]++;
     }
     rep(i,,n2)
     {
         b[i]=lower_bound(c+,c+t+,b[i])-c;
         d[b[i]][]++;
     }

     int m=;
     s[]=,s[]=;
     rep(i,,t)
     {
         if(d[i][]&&d[i][])
         {
             m++;
             q[m].x=d[i][];
             q[m].y=d[i][];
             continue;
         }
         if(d[i][]) s[]+=d[i][];
         if(d[i][]) s[]+=d[i][];
     }
     sort(q+,q+m+,cmp);
     int v=;
     rep(i,,m)
     {
         v^=;
         if(v==) s[v]+=q[i].x;
          else s[v]+=q[i].y;
     }
     if(s[]>s[]) printf("Cuber QQ\n");
      else printf("Quber CC\n");
     rep(i,,t)
      rep(j,,) d[i][j]=;
 }

 int main()
 {
     int cas=read();
     while(cas--) solve();
     return ;
 }