【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

  1 #include<bits/stdc++.h>
  2 using namespace std;
  3 typedef long long ll;
  4 typedef unsigned int uint;
  5 typedef unsigned long long ull;
  6 typedef pair<int,int> PII;
  7 typedef pair<ll,ll> Pll;
  8 typedef vector<int> VI;
  9 typedef vector<PII> VII;
 10 //typedef pair<ll,ll>P;
 11 #define N  200010
 12 #define M  200010
 13 #define fi first
 14 #define se second
 15 #define MP make_pair
 16 #define pb push_back
 17 #define pi acos(-1)
 18 #define mem(a,b) memset(a,b,sizeof(a))
 19 #define rep(i,a,b) for(int i=(int)a;i<=(int)b;i++)
 20 #define per(i,a,b) for(int i=(int)a;i>=(int)b;i--)
 21 #define lowbit(x) x&(-x)
 22 #define Rand (rand()*(1<<16)+rand())
 23 #define id(x) ((x)<=B?(x):m-n/(x)+1)
 24 #define ls p<<1
 25 #define rs p<<1|1
 26 
 27 const int MOD=1e9+7,inv2=(MOD+1)/2;
 28       double eps=1e-4;
 29       int INF=1e9;
 30       int inf=0x7fffffff;
 31       int dx[4]={-1,1,0,0};
 32       int dy[4]={0,0,-1,1};
 33 
 34 struct node
 35 {
 36     int x,y;
 37 }q[N];
 38 
 39 int a[N],b[N],c[N],s[2],d[N][2];
 40 ull k1,k2;
 41 
 42 int read()
 43 {
 44    int v=0,f=1;
 45    char c=getchar();
 46    while(c<48||57<c) {if(c=='-') f=-1; c=getchar();}
 47    while(48<=c&&c<=57) v=(v<<3)+v+v+c-48,c=getchar();
 48    return v*f;
 49 }
 50 
 51 unsigned long long rng()
 52 {
 53     unsigned long long k3 = k1, k4 = k2;
 54     k1 = k4;
 55     k3 ^= k3 << 23;
 56     k2 = k3 ^ k4 ^ (k3 >> 17) ^ (k4 >> 26);
 57     return k2 + k4;
 58 }
 59 
 60 bool cmp(node a,node b)
 61 {
 62     return a.x+a.y>b.x+b.y;
 63 }
 64 
 65 void solve()
 66 {
 67     int n1=read(),n2=read(),p=read();
 68     if(p==1)
 69     {
 70         rep(i,1,n1) a[i]=read();
 71         rep(i,1,n2) b[i]=read();
 72     }
 73      else
 74      {
 75          int mod;
 76          scanf("%llu%llu%d",&k1,&k2,&mod);
 77          rep(i,1,n1) a[i]=rng()%mod;
 78          scanf("%llu%llu%d",&k1,&k2,&mod);
 79          rep(i,1,n2) b[i]=rng()%mod;
 80      }
 81     int t=0;
 82     rep(i,1,n1) c[++t]=a[i];
 83     rep(i,1,n2) c[++t]=b[i];
 84     sort(c+1,c+t+1);
 85     t=unique(c+1,c+t+1)-c-1;
 86     rep(i,1,n1)
 87     {
 88         a[i]=lower_bound(c+1,c+t+1,a[i])-c;
 89         d[a[i]][0]++;
 90     }
 91     rep(i,1,n2)
 92     {
 93         b[i]=lower_bound(c+1,c+t+1,b[i])-c;
 94         d[b[i]][1]++;
 95     }
 96 
 97     int m=0;
 98     s[0]=0,s[1]=0;
 99     rep(i,1,t)
100     {
101         if(d[i][0]&&d[i][1])
102         {
103             m++;
104             q[m].x=d[i][0];
105             q[m].y=d[i][1];
106             continue;
107         }
108         if(d[i][0]) s[0]+=d[i][0];
109         if(d[i][1]) s[1]+=d[i][1];
110     }
111     sort(q+1,q+m+1,cmp);
112     int v=1;
113     rep(i,1,m)
114     {
115         v^=1;
116         if(v==0) s[v]+=q[i].x;
117          else s[v]+=q[i].y;
118     }
119     if(s[0]>s[1]) printf("Cuber QQ\n");
120      else printf("Quber CC\n");
121     rep(i,1,t)
122      rep(j,0,1) d[i][j]=0;
123 }
124 
125 
126 
127 int main()
128 {
129     int cas=read();
130     while(cas--) solve();
131     return 0;
132 }

 

posted on 2019-10-11 15:14  myx12345  阅读(181)  评论(0)  编辑  收藏  举报

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