A. Alien Rhyme
题意：
思路：将字符反向插入一颗Trie，然后自下而上的贪心即可，即先选后缀长的，再选后缀短的。
实现：
1 #include <iostream>
2 #include <cstdio>
3 #include <cstdlib>
4 #include <iomanip>
5
6 #include <vector>
7 #include <cstring>
8 #include <string>
9 #include <queue>
10 #include <deque>
11 #include <stack>
12 #include <map>
13 #include <set>
14
15 #include <utility>
16 #include <list>
17
18 #include <cmath>
19 #include <algorithm>
20 #include <cassert>
21 #include <bitset>
22 #include <complex>
23 #include <climits>
24 #include <functional>
25 #include <unordered_set>
26 #include <unordered_map>
27 using namespace std;
28
29 typedef long long ll;
30 typedef pair<int, int> ii;
31 typedef pair<ll, ll> l4;
32 typedef pair<double, double> dd;
33 #define mp make_pair
34 #define pb push_back
35
36 #define debug(x) cerr << #x << " = " << x << " "
37 const int maxlen = 50+1;
38 const int N = 1000+1;
39 const int maxn = N * maxlen;
40 const int A = 26;
41 int g[maxn][A], cnt[maxn], tot;
42 inline int newnode()
43 {
44 memset(g[tot], 0, sizeof(g[tot]));
45 cnt[tot] = 0;
46 return tot++;
47 }
48 void init()
49 {
50 tot = 0;
51 newnode();
52 }
53 char s[maxlen];
54 void insert()
55 {
56 int len = strlen(s);
57 int cur = 0;
58 for (int i = len-1; i >= 0; --i)
59 {
60 int c = s[i] - ‘A‘;
61 if (g[cur][c] == 0) g[cur][c] = newnode();
62 cur = g[cur][c];
63 ++cnt[cur];
64 }
65 }
66 int solve(int cur)
67 {
68 if (cnt[cur] == 1) return 0;
69 int ret = 0;
70 for (int c = 0; c < A; ++c)
71 if (g[cur][c])
72 {
73 //cerr << "from " << cur << " via " << char(‘A‘+c) << " to " << g[cur][c] << endl;
74 ret += solve(g[cur][c]);
75 }
76 int tmp = cnt[cur] - ret;
77 if (tmp >= 2) ret += 2;
78 return ret;
79 }
80 int main()
81 {
82 // ios::sync_with_stdio(false);
83 // cin.tie(0);
84 // int T; cin >> T;
85 int T; scanf("%d", &T);
86 for (int kase = 1; kase <= T; ++kase)
87 {
88 int n; scanf("%d", &n);
89 init();
90 for (int i = 0; i < n; ++i)
91 {
92 scanf("%s", s);
93 insert();
94 }
95 printf("Case #%d: %d\n", kase, solve(0));
96 }
97 }
98 /*
99 1
100 10 5 5
101 101010101000010100101
102 0 2 4 6 8 13 15 18 20
103 */
B. Golf Gophers
题意：
思路：设所求的人数为TARGET。我们可以把18个计数器弱化为1个计数器，具体就是每个计数器的值都设为同一个，计作MOD，这样我们每次询问通过求各计数器之和得到TARGET%MOD的值。这个看起来是不是很像中国剩余定理？发现小于18的质数刚好7个{2,3,5,7,11,13,17}，而且每个质数的最高幂{16,9,5,7,11,13,17}的乘积大于1e6（TARGET的最大可能值），所以就迎刃而解了
实现：
1 #include <iostream>
2 #include <cstdio>
3 #include <cstdlib>
4 #include <iomanip>
5
6 #include <vector>
7 #include <cstring>
8 #include <string>
9 #include <queue>
10 #include <deque>
11 #include <stack>
12 #include <map>
13 #include <set>
14
15 #include <utility>
16 #include <list>
17
18 #include <cmath>
19 #include <algorithm>
20 #include <cassert>
21 #include <bitset>
22 #include <complex>
23 #include <climits>
24 #include <functional>
25 #include <unordered_set>
26 #include <unordered_map>
27 using namespace std;
28
29 typedef long long ll;
30 typedef pair<int, int> ii;
31 typedef pair<ll, ll> l4;
32 typedef pair<double, double> dd;
33 #define mp make_pair
34 #define pb push_back
35
36 #define debug(x) cerr << #x << " = " << x << " "
37
38
39 int T, N, M;
40 vector<int> query(int mycnt, int myvalue, int judgecnt)
41 {
42 for (int i = 0; i < mycnt; ++i)
43 printf("%d%c", myvalue, " \n"[i+1==mycnt]);
44 fflush(stdout);
45 vector<int> ret(judgecnt, 0);
46 for (auto & e : ret)
47 scanf("%d", &e);
48 return ret;
49 }
50 vector<int> factor = {16, 9, 5, 7, 11, 13, 17};
51 vector<int> inv;
52 int all;
53 ll power(ll base, ll p, ll mod)
54 {
55 ll ret = 1 % mod;
56 while (p)
57 {
58 if (p&1) ret = ret * base % mod;
59 base = base * base % mod;
60 p >>= 1;
61 }
62 return ret;
63 }
64 ll inverse(ll base, ll mod)
65 {
66 return power(base, mod-2, mod);
67 }
68 void preprocess()
69 {
70 all = 1;
71 for (auto e : factor) all *= e;
72 for (auto e : factor)
73 {
74 int y = all/e;
75 ll _inv = -1;
76 for (int i = 1; i < e; ++i)
77 if (1ll * i * y % e == 1)
78 {
79 _inv = i;
80 break;
81 }
82 assert(_inv != -1);
83 inv.pb(y * _inv);
84 }
85 }
86 const int MILL = 18;
87 int process(int factor)
88 {
89 auto response = query(MILL, factor, MILL);
90 int ret = 0;
91 for (auto e : response) (ret += e) %= factor;
92 return factor;
93 }
94
95 void solve()
96 {
97 ll ret = 0;
98 for (int i = 0; i < factor.size(); ++i)
99 {
100 (ret += 1ll * process(factor[i]) * inv[i] % all) %= all;
101 }
102 auto tmp = query(1, ret, 1);
103 assert(tmp.size() == 1 && tmp.front() == 1);
104 }
105 int main()
106 {
107 preprocess();
108 scanf("%d %d %d", &T, &N, &M);
109 for (int kase = 1; kase <= T; ++kase)
110 solve();
111 }
C. Pylons
题意：
思路：比较明显的是在r或c比较大的时候，可以利用样例里的思路拓展一下（也许需要tweak一下奇偶性的问题）。r和c都比较小的时候可以利用和traveling salesman problem类似的方式打表（也可以暴搜）。
实现：写的太丑了就不贴了，而且我大约写了另外两个文件来生成数据/测试/打表。
原文地址：https://www.cnblogs.com/skyette/p/10700548.html
时间： 2024-09-29 08:41:37