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 }