题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=6655
题意是说两个人都有一些带有颜色的牌，两人轮流出牌，但是不能出对面出过的颜色的牌，最后谁不能出牌谁输。
贪心的想，如果这种牌我有对方也有，那我肯定先出以此来减少对方可出的牌，如果有几种这样的牌，那肯定是先出两方手牌总和最多的那种颜色。
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<string> 5 #include<algorithm> 6 #include<vector> 7 #include<map> 8 using namespace std; 9 typedef long long ll; 10 typedef unsigned long long ull; 11 const int maxn = 2e5 + 110; 12 struct node { 13 ll id, num; 14 node() {} 15 node(ll _id, ll _num) { id = _id, num = _num; } 16 }q[maxn]; 17 int a[maxn], b[maxn], num[maxn], numa[maxn], numb[maxn]; 18 ull k1, k2; 19 int mod; 20 ull rng() { 21 unsigned long long k3 = k1, k4 = k2; 22 k1 = k4; 23 k3 ^= k3 << 23; 24 k2 = k3 ^ k4 ^ (k3 >> 17) ^ (k4 >> 26); 25 return k2 + k4; 26 } 27 bool cmp(node a, node b) { 28 return a.num > b.num; 29 } 30 int main() { 31 int t; 32 scanf("%d", &t); 33 while (t--) { 34 int n, m, p, len = 0, cnt = 0; 35 scanf("%d%d%d", &n, &m, &p); 36 if (p == 1) { 37 for (int i = 1; i <= n; i++) 38 scanf("%d", &a[i]), num[++len] = a[i]; 39 for (int i = 1; i <= m; i++) 40 scanf("%d", &b[i]), num[++len] = b[i]; 41 } 42 else { 43 scanf("%llu%llu%d", &k1, &k2, &mod); 44 for (int i = 1; i <= n; i++) 45 a[i] = (int)(rng() % mod), num[++len] = a[i]; 46 scanf("%llu%llu%d", &k1, &k2, &mod); 47 for (int i = 1; i <= m; i++) 48 b[i] = (int)(rng() % mod), num[++len] = b[i]; 49 } 50 sort(num + 1, num + 1 + len); 51 len = unique(num + 1, num + 1 + len) - num - 1; 52 for (int i = 0; i <= len; i++) 53 numa[i] = numb[i] = 0; 54 for (int i = 1; i <= n; i++) { 55 int x = lower_bound(num + 1, num + 1 + len, a[i]) - num; 56 numa[x]++; 57 } 58 for (int i = 1; i <= m; i++) { 59 int x = lower_bound(num + 1, num + 1 + len, b[i]) - num; 60 numb[x]++; 61 } 62 for (int i = 1; i <= len; i++) { 63 if (numa[i] && numb[i]) 64 q[++cnt] = node(i, numa[i] + numb[i]); 65 } 66 sort(q + 1, q + 1 + cnt, cmp); 67 int w = 1, suma = n, sumb = m; 68 for (int i = 1; i <= cnt; i++) { 69 if (w & 1) 70 suma--, sumb -= numb[q[i].id]; 71 else 72 sumb--, suma -= numa[q[i].id]; 73 w++; 74 } 75 if (w & 1) { 76 if (suma > sumb) 77 printf("Cuber QQ "); 78 else 79 printf("Quber CC "); 80 81 } 82 else { 83 if (sumb > suma) 84 printf("Quber CC "); 85 else 86 printf("Cuber QQ "); 87 88 } 89 } 90 }