Codeforces Round #457 (Div. 2)

A.模拟。

B.模拟。比赛的时候这题出锅了，然后就unrated了。注意要先保证最大值最小，再保证字典序最大。

C.构造。挑一个大于N的质数然后连啊连就可以了。

D.可持久化Trie。弄两个Trie。一个记录字符串，另一个记录数字。

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<iostream>
 4 using namespace std;
 5 struct StringTrie {
 6     StringTrie *ch[26]; int val;
 7     StringTrie() {memset(this, 0, sizeof(*this)); val = -1;}
 8     StringTrie* set(int n, const char *str, int p, int v) {
 9         StringTrie *res = new StringTrie(); *res = *this;
10         if (p >= n) res->val = v;
11         else {
12             int c = str[p] - ‘a‘;
13             if (!ch[c]) ch[c] = new StringTrie();
14             res->ch[c] = ch[c]->set(n, str, p + 1, v);
15         } return res;
16     }
17     int query(int n, const char *str, int p) {
18         if (p >= n) return val;
19         else {
20             int c = str[p] - ‘a‘;
21             if (!ch[c]) return -1;
22             return ch[c]->query(n, str, p + 1);
23         }
24     }
25 } *srt[100005];
26 struct BinaryTrie {
27     BinaryTrie *ch[2]; int val;
28     BinaryTrie() {memset(this, 0, sizeof(*this));}
29     BinaryTrie* add(int x, int val, int p) {
30         BinaryTrie *res = new BinaryTrie(); *res = *this; res->val += val;
31         if (p >= 0) {
32             int c = (x >> p) & 1;
33             if (!ch[c]) ch[c] = new BinaryTrie();
34             res->ch[c] = ch[c]->add(x, val, p - 1);
35         } return res;
36     }
37     int query(int x, int p) {
38         if (p < 0) return 0;
39         else {
40             int c = (x >> p) & 1, res = 0;
41             if (c && ch[0]) res += ch[0]->val;
42             if (ch[c]) res += ch[c]->query(x, p - 1);
43             return res;
44         }
45     }
46 } *brt[100005];
47 int main() {
48     srt[0] = new StringTrie(); brt[0] = new BinaryTrie();
49     int n; scanf("%d", &n); char op[20], str[20];
50     for (int v, i = 1; i <= n; ++i) {
51         scanf("%s", op);
52         if (*op == ‘s‘) {
53             scanf("%s%d", str, &v); int l = strlen(str);
54             int x = srt[i-1]->query(l, str, 0);
55             srt[i] = srt[i-1]->set(l, str, 0, v);
56             brt[i] = brt[i-1];
57             if (~x) brt[i] = brt[i]->add(x, -1, 30);
58             brt[i] = brt[i]->add(v, 1, 30);
59         } else if (*op == ‘r‘) {
60             scanf("%s", str); int l = strlen(str);
61             int x = srt[i-1]->query(l, str, 0);
62             brt[i] = brt[i-1]; srt[i] = srt[i-1]->set(l, str, 0, -1);
63             if (~x) brt[i] = brt[i]->add(x, -1, 30);
64         } else if (*op == ‘q‘) {
65             scanf("%s", str); int l = strlen(str);
66             brt[i] = brt[i-1]; srt[i] = srt[i-1];
67             int x = srt[i]->query(l, str, 0);
68             if (~x) printf("%d\n", brt[i]->query(x, 30));
69             else puts("-1"); fflush(stdout);
70         } else {
71             scanf("%d", &v); srt[i] = srt[i-v-1]; brt[i] = brt[i-v-1];
72         }
73     }
74     return 0;
75 }

E.跟bzoj上的遥远的国度很类似。

原文地址：https://www.cnblogs.com/p0ny/p/8328207.html