A：SUM
水。
1 #include<bits/stdc++.h>
2
3 using namespace std;
4
5 #define N 100010
6 typedef long long ll;
7
8 int n;
9 ll arr[N];
10 ll sum[N];
11
12 int main()
13 {
14 while(~scanf("%d",&n))
15 {
16 memset(sum, 0, sizeof sum);
17 for(int i = 1; i <= n; ++i)
18 {
19 scanf("%lld",&arr[i]);
20 }
21 for(int i = n; i >= 1; --i)
22 {
23 sum[i] = sum[i + 1] + arr[i] * (n - i + 1);
24 }
25 for(int i = 1;i <= n; ++i)
26 {
27 printf("%lld%c",sum[i]," \n"[i == n]);
28 }
29 }
30 return 0;
31 }
B：XOR1
思路：枚举ai, 去字典树中找最大值，取max
1 #include <bits/stdc++.h>
2
3 using namespace std;
4
5 #define N 100010
6 #define ll long long
7
8 int tree[N * 64][5];
9 int cnt[N * 64];
10 int pos;
11
12 inline void Init()
13 {
14 memset(tree, 0, sizeof tree);
15 memset(cnt, 0, sizeof cnt);
16 pos = 0;
17 }
18
19 inline void insert(int x)
20 {
21 bitset <40> b; b = x;
22 int root = 0;
23 for (int i = 39; i >= 0; --i)
24 {
25 int id = b[i];
26 if (!tree[root][id])
27 tree[root][id] = ++pos;
28 root = tree[root][id];
29 cnt[root]++;
30 }
31 }
32
33 inline ll Find(int x)
34 {
35 bitset <40> b; b = x;
36 ll ans = 0;
37 int root = 0;
38 for (int i = 39; i >= 0; --i)
39 {
40 int id = b[i] ^ 1;
41 bool flag = true;
42 if (!tree[root][id] || cnt[tree[root][id]] <= 0)
43 id ^= 1, flag = false;
44 root = tree[root][id];
45 if (flag) ans += (1 << i);
46 }
47 return ans;
48 }
49
50 int n;
51 int arr[N];
52
53 int main()
54 {
55 while (scanf("%d", &n) != EOF)
56 {
57 Init();
58 for (int i = 1; i <= n; ++i)
59 {
60 scanf("%d", arr + i);
61 insert(arr[i]);
62 }
63 ll ans = 0;
64 for (int i = 1; i < n; ++i)
65 {
66 ans = max(ans, Find(arr[i]));
67 }
68 printf("%lld\n", ans);
69 }
70 return 0;
71 }
C：XOR2
思路：先DFS跑出DFS序，然后配合可持久化线段树就可以对子树进行区间操作。
我们自底向上推，这样对于一棵树，它的孩子的答案都已经更新，那么先取max, 假设直系孩子有n个，那么只需要枚举n - 1个孩子以及它的子树里的点更新答案就可以。
1 #include <bits/stdc++.h>
2
3 using namespace std;
4
5 #define N 100010
6 #define ll long long
7
8 int arr[N];
9
10 struct Edge
11 {
12 int to, nx;
13 inline Edge() {}
14 inline Edge(int to, int nx) : to(to), nx(nx) {}
15 }edge[N << 1];
16
17 int head[N], pos, cnt;
18
19 inline void Init()
20 {
21 memset(head, -1, sizeof head);
22 pos = 0; cnt = 0;
23 }
24
25 inline void addedge(int u, int v)
26 {
27 edge[++cnt] = Edge(v, head[u]); head[u] = cnt;
28 edge[++cnt] = Edge(u, head[v]); head[v] = cnt;
29 }
30
31 struct Point
32 {
33 int fa, ord, son, id;
34 int ans;
35 inline bool operator < (const Point& r) const
36 {
37 return son - ord < r.son - r.ord;
38 }
39 }point[N];
40
41 int n;
42 int ford[N];
43
44 struct node
45 {
46 int son[2], cnt;
47 inline node()
48 {
49 memset(son, 0, sizeof son);
50 cnt = 0;
51 }
52 }tree[N * 64];
53
54 int root[N];
55 int tot;
56
57 inline void insert(int id, int x)
58 {
59 root[id] = ++tot;
60 int pre = root[id - 1];
61 int now = root[id];
62 tree[now] = tree[pre];
63 bitset <32> b; b = x;
64 for (int i = 31; i >= 0; --i)
65 {
66 int index = b[i];
67 tree[++tot] = tree[tree[now].son[index]];
68 tree[tot].cnt++;
69 tree[now].son[index] = tot;
70 now = tot;
71 }
72 }
73
74 inline int Find(int l, int r, int x)
75 {
76 int ans = 0;
77 l = root[l], r = root[r];
78 bitset <32> b; b = x;
79 for (int i = 31; i >= 0; --i)
80 {
81 int index = b[i] ^ 1;
82 bool flag = true;
83 if (tree[tree[r].son[index]].cnt - tree[tree[l].son[index]].cnt <= 0)
84 {
85 index ^= 1;
86 flag = false;
87 }
88 if (flag) ans += (1 << i);
89 r = tree[r].son[index]; l = tree[l].son[index];
90 }
91 return ans;
92 }
93
94 struct Node
95 {
96 int id, cnt;
97 inline Node() {}
98 inline Node(int id, int cnt) : id(id), cnt(cnt) {}
99 inline bool operator < (const Node &r) const
100 {
101 return cnt < r.cnt;
102 }
103 };
104
105 inline void DFS(int u)
106 {
107 point[u].ord = ++pos;
108 point[u].id = u;
109 point[u].ans = 0;
110 insert(pos, arr[u]);
111 ford[pos] = u;
112 vector <Node> vv;
113 for (int it = head[u]; ~it; it = edge[it].nx)
114 {
115 int v = edge[it].to;
116 if (v == point[u].fa) continue;
117 point[v].fa = u; DFS(v);
118 point[u].ans = max(point[u].ans, point[v].ans);
119 vv.emplace_back(v, point[v].son - point[v].ord);
120 }
121 point[u].son = pos;
122 int l = point[u].ord, r = point[u].son;
123 if (l == r) return;
124 if (l + 1 == r)
125 {
126 point[u].ans = arr[ford[l]] ^ arr[ford[r]];
127 return;
128 }
129 point[u].ans = max(point[u].ans, Find(l - 1, r, arr[u]));
130 sort(vv.begin(), vv.end());
131 for (int i = 0, len = vv.size(); i < len - 1; ++i)
132 {
133 int it = vv[i].id;
134 int L = point[it].ord, R = point[it].son;
135 for (int j = L; j <= R; ++j)
136 {
137 point[u].ans = max(point[u].ans, Find(l - 1, r, arr[ford[j]]));
138 }
139 }
140 }
141
142 int main()
143 {
144 while (scanf("%d", &n) != EOF)
145 {
146 Init();
147 memset(root, 0, sizeof root);
148 tot = 0;
149 for (int i = 1; i <= n; ++i)
150 scanf("%d", arr + i);
151 for (int i = 1, u, v; i < n; ++i)
152 {
153 scanf("%d%d", &u, &v);
154 addedge(u, v);
155 }
156 DFS(1);
157 for (int i = 1; i <= n; ++i) printf("%d%c", point[i].ans, " \n"[i == n]);
158 }
159 return 0;
160 }
D：String
留坑。
E：Dirt
留坑。
F：Poker
水。
1 #include <bits/stdc++.h>
2
3 using namespace std;
4
5 #define N 100010
6
7 int n;
8 int arr[N];
9 int brr[N];
10 int a[N];
11
12 inline void Init()
13 {
14 memset(a, 0, sizeof a);
15 }
16
17 inline int lowbit(int x)
18 {
19 return x & (-x);
20 }
21
22 inline void update(int x, int val)
23 {
24 for (int i = x; i <= n; i += lowbit(i))
25 a[i] += val;
26 }
27
28 inline int sum(int x)
29 {
30 int ans = 0;
31 for (int i = x; i > 0; i -= lowbit(i))
32 ans += a[i];
33 return ans;
34 }
35
36 inline bool check(int mid, int emp)
37 {
38 int tot = sum(mid);
39 return tot <= emp;
40 }
41
42 int main()
43 {
44 while (scanf("%d", &n) != EOF)
45 {
46 Init();
47 for (int i = 1; i <= n; ++i) update(i, 1);
48 for(int i = 1; i <= n; ++i)
49 {
50 scanf("%d",&arr[i]);
51 }
52 for (int i = 1; i <= n; ++i)
53 {
54 int index = (int)floor(sqrt(n - i + 1));
55 int l = 1, r = n, id;
56 while (r - l >= 0)
57 {
58 int mid = (l + r) >> 1;
59 int tot = sum(mid);
60 if (tot == index)
61 id = mid;
62 if (tot >= index)
63 r = mid - 1;
64 else
65 l = mid + 1;
66 }
67 brr[id] = arr[i];
68 update(id, -1);
69 }
70 for (int i = 1; i <= n; ++i) printf("%d%c", brr[i], " \n"[i == n]);
71 }
72 return 0;
73 }
1 #include<bits/stdc++.h>
2
3 using namespace std;
4
5 #define N 100010
6
7 int n;
8 int arr[N];
9 int brr[N];
10
11 int main()
12 {
13 while(~scanf("%d",&n))
14 {
15 for(int i = 1; i <= n; ++i)
16 {
17 scanf("%d",&arr[i]);
18 }
19 vector<int>vec;
20 for(int i = n; i >= 1;--i)
21 {
22 int tmp = floor(sqrt(n - i + 1));
23 vec.insert(vec.begin() + tmp - 1, arr[i]);
24 }
25 for(int i = 0; i < n; ++i)
26 {
27 printf("%d%c",vec[i]," \n"[i == n - 1]);
28 }
29 }
30 return 0;
31 }
原文地址：https://www.cnblogs.com/Dup4/p/9477456.html
时间： 2024-10-20 04:12:42