谢谢大家参与本次比赛~！
难度分配
简单题：ABCFH
中等题：DE
难题：GIJ
A
直接输入四个字符串，直接判断输出即可。无坑点
#include<iostream>#include<string>using namespace std;int main(){string a,b,c,d;while(cin>>a>>b>>c>>d){d.pop_back();cout<<"I can "<<c<<" "<<d<<"!"<<endl;}return 0;}
B
用map实现每一个操作即可，具体看代码。如果不会map可以把每一个字符串hash成数字。
有一个坑点，就是在还没教会机器人的时候，问题就已经被禁止回答了，所以要用多个map去标记。
#include <iostream>#include <string>#include <map>using namespace std;int main(){map<string, string> mp;map<string, int> vis;string a, b;int Q;scanf("%d", &Q);while (Q--){int t;scanf("%d", &t);if (t == 1){cin >> a >> b;mp[a] = b;if (vis[a] == 0)vis[a] = 1;}if (t == 2){cin >> a;if (mp[a] != "" && vis[a] == 1){cout << mp[a] << endl;}else{cout << "I don't know what to say!" << endl;}}if (t == 3){cin >> a;vis[a] = 2;}if (t == 4){cin >> a;vis[a] = 1;}}return 0;}
C
多写几项找规律得到公式
#include <iostream>using namespace std;typedef long long ll;int main(){ll N;while (cin >> N){if (N % 2 == 0)cout << N * N + N / 2 << endl;elsecout << N * N << endl;}return 0;}
D
数据范围小，直接上Floyd求传递闭包。很经典的一道题目。离散数学有教。注意循环的顺序。
#include <iostream>using namespace std;typedef long long ll;bool G[305][305];int main(){int N;scanf("%d", &N);for (int i = 1; i <= N; i++)for (int j = 1; j <= N; j++)scanf("%d", &G[i][j]);for (int k = 1; k <= N; ++k){for (int i = 1; i <= N; ++i){for (int j = 1; j <= N; ++j){if (G[i][j] == 1 || (G[i][k] == 1 && G[k][j] == 1)){G[i][j] = 1;}}}}int Q;scanf("%d", &Q);while (Q--){int u, v;scanf("%d%d", &u, &v);printf("%d\n", G[u][v]);}return 0;}
E
很经典的线段树维护最大值，没有任何坑点。这里没有必要开四个线段树，具体看代码实现！
#include <iostream>#include <string.h>#include <map>using namespace std;#define mod 10000007typedef long long ll;const int MAXN = 100005;ll tree[MAXN << 2];ll LI[4][MAXN];void pushup(int rt){tree[rt] = max(tree[rt << 1], tree[rt << 1 | 1]);}void update(int K, int C, int X, int l, int r, int rt){if (l == r){LI[K][C] = X;tree[rt] = LI[0][C] * LI[1][C] * LI[2][C] * LI[3][C];return;}int m = (l + r) / 2;if (C <= m)update(K, C, X, l, m, rt << 1);elseupdate(K, C, X, m + 1, r, rt << 1 | 1);pushup(rt);}ll query(int L, int R, int l, int r, int rt){if (L <= l && r <= R){return tree[rt];}ll ans = 0x3f3f3f3f3f3f3f3f;ans = -ans;int m = (l + r) / 2;if (L <= m)ans = max(ans, query(L, R, l, m, rt << 1));if (R > m)ans = max(ans, query(L, R, m + 1, r, rt << 1 | 1));return ans;}int main(){int N, Q;cin >> N >> Q;while (Q--){int t, K, C, X, L, R;cin >> t;if (t == 1){cin >> K >> C >> X;update(K - 1, C, X, 1, N, 1);}else{cin >> L >> R;cout << query(L, R, 1, N, 1) << endl;}}return 0;}
F
暴力枚举即可，提取每一位，看看有没有重复数字，没有重复就输出。
#include <iostream>#include <cstring>using namespace std;bool vis[10];int num[3];void solve(int n){if (n == 3 && num[0] * 2 == num[1] && num[0] * 3 == num[2])cout << num[0] << ' ' << num[1] << ' ' << num[2] << endl;for (int i = 1; i <= 9; i++){if (!vis[i]){vis[i] = true;for (int j = 1; j <= 9; j++){if (!vis[j]){vis[j] = true;for (int k = 1; k <= 9; k++){if (!vis[k]){vis[k] = true;num[n] = i * 100 + j * 10 + k;solve(n + 1);vis[k] = false;}}vis[j] = false;}}vis[i] = false;}}}int main(){string a;cin >> a;memset(vis, false, sizeof(vis));solve(0);}
G
很经典的一道题目。考察二分+最小生成树
我们考虑将所有的白色的边加上一个cost，如x至y有一条白色边权值为len，那我们将它的权值变为cost+len，然后我们做最小生成树，很明显，如果加上的cost越大，所选的白色边越少，所以我们就二分这个cost，直至刚好选了need条时，则答案就是生成树的权值-need*cost。
#include <iostream>#include <set>#include <map>#include <cstdio>#include <cstring>#include <cstdlib>#include <ctime>#include <vector>#include <queue>#include <algorithm>#include <cmath>#define inf 1000000000#define pa pair<int, int>#define ll long longusing namespace std;int read(){int x = 0, f = 1;char ch = getchar();while (ch < '0' || ch > '9'){if (ch == '-')f = -1;ch = getchar();}while (ch >= '0' && ch <= '9'){x = x * 10 + ch - '0';ch = getchar();}return x * f;}int n, m, cnt, tot, ned;int sumv;int u[100005], v[100005], w[100005], c[100005];int fa[100005];struct edge{int u, v, w, c;} e[100005];bool operator<(edge a, edge b){return a.w == b.w ? a.c < b.c : a.w < b.w;}int find(int x){return x == fa[x] ? x : fa[x] = find(fa[x]);}bool check(int x){tot = cnt = 0;for (int i = 1; i <= n; i++)fa[i] = i;for (int i = 1; i <= m; i++){e[i].u = u[i], e[i].v = v[i], e[i].w = w[i];e[i].c = c[i];if (!c[i])e[i].w += x;}sort(e + 1, e + m + 1);for (int i = 1; i <= m; i++){int p = find(e[i].u), q = find(e[i].v);if (p != q){fa[p] = q;tot += e[i].w;if (!e[i].c)cnt++;}}return cnt >= ned;}int main(){n = read();m = read();ned = read();for (int i = 1; i <= m; i++){u[i] = read(), v[i] = read(), w[i] = read(), c[i] = read();u[i]++;v[i]++;}int l = -105, r = 105;while (l <= r){int mid = (l + r) >> 1;if (check(mid))l = mid + 1, sumv = tot - ned * mid;elser = mid - 1;}printf("%d", sumv);return 0;}
H
模拟题。暴力枚举端点即可，但是不能用 N^3 的复杂度。第二个端点其实不用枚举，直接取中点即可，优化到了 N^2.
注意A 不能和B相等。像aaa这种输入就得输出SAD而不是a a a。
#include <iostream>#include <string.h>#include <map>#include <vector>using namespace std;#define mod 10000007typedef long long ll;char str[505];int main(){int T;scanf("%d", &T);while (T--){scanf("%s", str);int len = strlen(str);bool yes = 1;for (int i = 1; i <= len - 2; i++){if ((len - i) % 2 == 1)continue;string a, b, c;for (int k = 0; k < i; k++)a.push_back(str[k]);for (int k = i; k < i + (len - i) / 2; k++)b.push_back(str[k]);for (int k = i + (len - i) / 2; k < len; k++)c.push_back(str[k]);if (a != b && b == c){cout << a << " " << b << " " << c << endl;yes = 0;break;}}if (yes)cout << "SAD" << endl;}return 0;}
I
数学题
要推出任意次方和公式
我们会发现，求k次和的时候，需要用到k-1,k-2…,0的结果。
所以实际求的时候，可以反过来逐个计算。
组合数我们可以事先计算好，除以k+1，也可以预处理乘法逆元。
所以总的复杂度O(k^2)。
#include <bits/stdc++.h>using namespace std;typedef long long LL;const int mod = 1e9 + 7;const int N = 2050;int T, y;LL x, C[N][N], rev[N], dp[N];LL pow_mod(LL x, LL p){LL s = 1;while (p){if (p & 1)s = s * x % mod;x = x * x % mod;p >>= 1;}return s;}void init(){rev[0] = 1;C[0][0] = 1;for (int i = 1; i < N; i++){C[i][0] = C[i][i] = 1;for (int j = 1; j < i; j++){C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % mod;}rev[i] = pow_mod(i, mod - 2);}}LL solve1(LL x, int y){x = x % mod;dp[0] = x;LL n = (x + 1) % mod;for (int i = 1; i <= y; i++){dp[i] = (pow_mod(n, i + 1) + mod - 1) % mod;for (int j = 1; j <= i; j++){dp[i] = (dp[i] - C[i + 1][j + 1] * dp[i - j] % mod) % mod;}if (dp[i] < 0)dp[i] += mod;dp[i] = dp[i] * rev[i + 1] % mod;}return dp[y];}LL solve2(int x, LL y){LL ret = 0;for (int i = 1; i <= x; i++){ret = (ret + pow_mod(i, y)) % mod;}return ret;}int main(){init();scanf("%d", &T);while (T--){scanf("%lld %d", &x, &y);printf("%lld\n", (solve1(x, y) + solve2(y, x)) % mod);}return 0;}
J
一道防AK题
做法是 主席树+字典树+二分
我们先把所有ADD的字符串插入到一颗字典树。实际上每一次查询，就是查询字典树某个子树中某一层的节点的权值和。
首先我们可以用dfs序给字典树每一个节点编号记为dfn，然后再做一次宽搜，给每一个节点再编一次编号记为bfn。
然后我们记录字典树每一层的节点都有哪些。
然后我们再用把树拍扁的思想，记录每一个节点所代表的区间。然后建一棵线段树。
然后对于ADD操作，我们直接找到那个字符串在字典树中的最后一个节点，对那个bfn所在地方+1，用线段树维护和。
然后对于QUERY操作，我们直接找到那个前缀所在的节点的那个子树，然后要找到，所要查询那个长度所在的层的哪些编号属于那个子树。
对于这个需求，我们直接对那一层的节点二分，二分值是节点所代表的的区间。这样子我们就能确定那一层哪些节点属于那个子树了。
对于回退问题，我们直接用可持久化线段树解决
额外的的解释：
全部读入，构建字典树，记录下每个ADD和QUERY的字符串末尾对应的节点ID；
对字典树做dfs
记录dfs序，dfs_lft[x]和dfs_rgt[x]分别表示以x为根的子树中的最小dfs序和最大dfs序；
将深度相同的节点放到一个数组，按dfs序升序排序，同时记录对应的节点ID；
按照节点深度和dfs序重新排列，即第一层所有节点的dfs序，第二层所有节点的dfs序…
同时记录每个节点跟这个新序列的映射关系idx[x]；
对这个序列构建主席树；
对于ADD操作，主席树开新版本，找到对应的idx进行加1；
对于REVERT操作，直接将当前版本指向指定版本；
对于QUERY操作，先找到对应的子树根节点x，根据dfs_lft[x]和dfs_rgt[x]，二分出在相应深度的dfs序，因为在x的子节点的dfs序必定大于等于dfs_lft[x]，小于等于dfs_rgt[x]，而非其子节点必定不满足此条件。根据这个dfs序范围反查出两端节点ID，进而转化为主席树的区间端点，然后进行简单的区间求和操作即可。
标程1：
#include <bits/stdc++.h>using namespace std;const int N = 1e5 + 10;int T, Q;char op[10], ss[N];int ch[N][26], oper[N], num[N], dep[N], node_id[N];int sz, max_dep, dfs_cnt, dfs_lft[N], dfs_rgt[N], idx[N];vector<int> V[N], G[N];int n, node_cnt, tree[N], S[N * 20], lson[N * 20], rson[N * 20];struct Trie{void init(){sz = 1;memset(ch[0], 0, sizeof(ch[0]));}int add(){int len = strlen(ss);int u = 0, c;for (int i = 0; i < len; i++){c = ss[i] - 'a';if (!ch[u][c]){memset(ch[sz], 0, sizeof(ch[sz]));ch[u][c] = sz++;}u = ch[u][c];}return u;}void dfs(int u){max_dep = max(max_dep, dep[u]);dfs_lft[u] = ++dfs_cnt;V[dep[u]].push_back(dfs_cnt);G[dep[u]].push_back(u);for (int i = 0; i < 26; i++){if (ch[u][i]){dep[ch[u][i]] = dep[u] + 1;dfs(ch[u][i]);}}dfs_rgt[u] = dfs_cnt;}} tr;int build(int L, int R){int ret = ++node_cnt;S[ret] = 0;if (L < R){int mid = L + R >> 1;lson[ret] = build(L, mid);rson[ret] = build(mid + 1, R);}else{lson[ret] = rson[ret] = 0;}return ret;}int add_node(int root0, int p, int v){int L = 1, R = n, mid;int root = ++node_cnt;int ret = root;S[root] = S[root0] + v;while (L < R){mid = L + R >> 1;if (p <= mid){lson[root] = ++node_cnt;rson[root] = rson[root0];root = lson[root];root0 = lson[root0];R = mid;}else{lson[root] = lson[root0];rson[root] = ++node_cnt;root = rson[root];root0 = rson[root0];L = mid + 1;}S[root] = S[root0] + v;}return ret;}int sum(int root, int L, int R, int ll, int rr){if (L == ll && R == rr){return S[root];}int mid = L + R >> 1;if (rr <= mid)return sum(lson[root], L, mid, ll, rr);if (ll > mid)return sum(rson[root], mid + 1, R, ll, rr);return sum(lson[root], L, mid, ll, mid) + sum(rson[root], mid + 1, R, mid + 1, rr);}int query(int root, int node, int len){if (len > max_dep || dep[node] > len)return 0;if (V[len].size() == 0)return 0;int lft = dfs_lft[node];int rgt = dfs_rgt[node];if (rgt < V[len][0] || lft > V[len].back())return 0;int low, top, mid, key_l, key_r;low = 0;top = V[len].size() - 1;key_l = top;while (low <= top){mid = low + top >> 1;if (V[len][mid] >= lft){key_l = min(key_l, mid);top = mid - 1;}else{low = mid + 1;}}low = 0;top = V[len].size() - 1;key_r = 0;while (low <= top){mid = low + top >> 1;if (V[len][mid] <= rgt){key_r = max(key_r, mid);low = mid + 1;}else{top = mid - 1;}}if (key_l > key_r)return 0;int idx_lft = idx[G[len][key_l]];int idx_rgt = idx[G[len][key_r]];return sum(root, 1, n, idx_lft, idx_rgt);}void init(){dfs_cnt = 0;max_dep = 0;dep[0] = 0;tr.dfs(0);n = 0;for (int i = 1; i <= max_dep; i++){for (int j = 0; j < G[i].size(); j++){idx[G[i][j]] = ++n;}}node_cnt = 0;tree[0] = build(1, n);}int main(){scanf("%d", &Q);tr.init();for (int i = 1; i <= Q; i++){scanf("%s", op);if (op[0] == 'A'){scanf("%s", ss);node_id[i] = tr.add();oper[i] = 0;}else if (op[0] == 'Q'){scanf("%s %d", ss, num + i);node_id[i] = tr.add();oper[i] = 1;}else{scanf("%d", num + i);oper[i] = 2;}}init();for (int i = 1; i <= Q; i++){if (oper[i] == 0){int j = idx[node_id[i]];tree[i] = add_node(tree[i - 1], j, 1);}else if (oper[i] == 1){tree[i] = tree[i - 1];printf("%d\n", query(tree[i], node_id[i], num[i]));}else{tree[i] = tree[num[i] - 1];}}for (int i = 0; i <= max_dep; i++){V[i].clear();G[i].clear();}return 0;}
标程2：
#include <iostream>#include <string.h>#include <map>#include <vector>#include <queue>#include <map>#include <algorithm>#include <stdio.h>using namespace std;#define MAXN 101005typedef long long ll;int nxt[MAXN][28];int cnt = 0;void insert(string str){int len = str.size();int cur = 0;for (int i = 0; i < len; i++){if (nxt[cur][str[i] - 'a'] == 0)nxt[cur][str[i] - 'a'] = ++cnt;cur = nxt[cur][str[i] - 'a'];}}int getid(string str){int len = str.size();int cur = 0;for (int i = 0; i < len; i++){if (nxt[cur][str[i] - 'a'] == 0)return 0;cur = nxt[cur][str[i] - 'a'];}return cur;}int ldfn[MAXN];int rdfn[MAXN];int mytime = 0;void dfs(int cur){ldfn[cur] = ++mytime;for (int i = 0; i < 26; i++){if (nxt[cur][i]){dfs(nxt[cur][i]);}}rdfn[cur] = mytime;}int bfn[MAXN];vector<int> ceng[MAXN * 2];int bt = 1;void bfs(){queue<pair<int, int>> que;que.push({0, 0});while (!que.empty()){pair<int, int> tp = que.front();que.pop();ceng[tp.first].push_back(tp.second);for (int i = 0; i < 27; i++){if (nxt[tp.second][i]){que.push({tp.first + 1, nxt[tp.second][i]});}}}}int root[MAXN];int mycount = 0;int tree[MAXN * 15];int rs[MAXN * 15];int ls[MAXN * 15];void update(int C, int X, int l, int r, int &rt, int lrt){rt = ++mycount;ls[rt] = ls[lrt];rs[rt] = rs[lrt];tree[rt] = tree[lrt] + X;if (l == r){return;}int m = (l + r) / 2;if (C <= m)update(C, X, l, m, ls[rt], ls[lrt]);elseupdate(C, X, m + 1, r, rs[rt], rs[lrt]);}int query(int L, int R, int l, int r, int rt){if (L <= l && r <= R){return tree[rt];}int m = (l + r) / 2;int ans = 0;if (L <= m)ans += query(L, R, l, m, ls[rt]);if (R > m)ans += query(L, R, m + 1, r, rs[rt]);return ans;}struct queryq{int type;string str;int len;} q[100505];char str[50];char tmp[100005];int main(){int Q;scanf("%d", &Q);for (int i = 1; i <= Q; i++){scanf("%s", str);if (str[0] == 'A'){scanf("%s", tmp);q[i].type = 0;q[i].str = tmp;q[i].len = 0;insert(tmp);}else if (str[0] == 'Q'){scanf("%s", tmp);int len;scanf("%d", &len);q[i].type = 1;q[i].str = tmp;q[i].len = len;}else{int len;scanf("%d", &len);q[i].type = 2;q[i].str = "";q[i].len = len;}}dfs(0);bfs();for (int i = 0; i <= 100005; i++){for (int j = 0; j < ceng[i].size(); j++){bfn[ldfn[ceng[i][j]]] = bt++;ceng[i][j] = ldfn[ceng[i][j]];}}cnt++;for (int i = 1; i <= Q; i++){if (q[i].type == 0){int tp = getid(q[i].str);update(bfn[ldfn[tp]], 1, 1, cnt, root[i], root[i - 1]);}else if (q[i].type == 1){root[i] = root[i - 1];int tp = getid(q[i].str);if (tp == 0){printf("0\n");continue;}if (q[i].len < q[i].str.size()){printf("0\n");continue;}if (q[i].len > 100005){printf("0\n");continue;}if (ceng[q[i].len].size() == 0){printf("0\n");continue;}if (ldfn[tp] > ceng[q[i].len].back()){printf("0\n");continue;}if (rdfn[tp] < ceng[q[i].len].front()){printf("0\n");continue;}int l;int r;if (ldfn[tp] <= ceng[q[i].len].front())l = ceng[q[i].len].front();elsel = *lower_bound(ceng[q[i].len].begin(), ceng[q[i].len].end(), ldfn[tp]);if (rdfn[tp] >= ceng[q[i].len].back())r = ceng[q[i].len].back();elser = *(upper_bound(ceng[q[i].len].begin(), ceng[q[i].len].end(), rdfn[tp]) - 1);printf("%d\n", query(bfn[l], bfn[r], 1, cnt, root[i]));}else{root[i] = root[q[i].len - 1];}}return 0;}
