开学前最后一场训练了，努力搞了个8题，还算可以。就是这场题没多少我能做的，只切了两道题。
题目链接：https://codeforces.com/gym/102428
D：
solver：czq
题意就是二维平面上有n颗恒星，每颗恒星有一个固定亮度。问：是否存在一条直线，从平面的一端扫到另一端，满足：对于任意两颗恒星S和T，如果S亮度大于T，那么S一定在T之前或者跟T一起同时被直线扫到。只需输出Y或N。
构造一下Y和N的例子你就会发现，对于所有亮度不同的恒星，从亮度高的恒星指向亮度低的恒星，这样我们就得到若干条向量。若答案为Y，那么必然存在一对向量，平行且方向相反，而且如果我们把所有的向量极角排序，这对平行的向量位置相邻。
1 /* basic header */ 2 #include <bits/stdc++.h> 3 /* define */ 4 #define ll long long 5 #define pb emplace_back 6 #define mp make_pair 7 #define eps 1e-8 8 #define lson (curpos<<1) 9 #define rson (curpos<<1|1) 10 /* namespace */ 11 using namespace std; 12 /* header end */ 13 14 const double pi = acos(-1); 15 const int maxn = 2020; 16 17 struct Star { 18 int x, y, b; 19 } star[maxn]; 20 vector<double>k; 21 int n; 22 23 double getRad(int x, int y) { 24 if (!y) return x > 0 ? 0 : pi; 25 if (!x) return y > 0 ? pi / 2 : -pi / 2; 26 double ans = atan(double(y) / x); 27 return x > 0 ? ans : ans + pi; 28 } 29 30 bool check() { 31 if (k.size() < 2) return true; 32 double last = k.back() - pi * 2; 33 for (auto x : k) { 34 if (pi - x + last <= eps) return true; 35 last = x; 36 } 37 return false; 38 } 39 40 int main() { 41 scanf("%d", &n); 42 for (int i = 0; i < n; i++) { 43 scanf("%d%d%d", &star[i].x, &star[i].y, &star[i].b); 44 for (int j = 0; j < i; j++) { 45 if (star[i].b > star[j].b) k.pb(getRad(star[i].x - star[j].x, star[i].y - star[j].y)); 46 else if (star[i].b < star[j].b) k.pb(getRad(star[j].x - star[i].x, star[j].y - star[i].y)); 47 } 48 } 49 sort(k.begin(), k.end()); 50 if (check()) puts("Y"); else puts("N"); 51 return 0; 52 }
E：
solver：lzh
1 #include <bits/stdc++.h> 2 using namespace std; 3 #define ff first 4 #define ss second 5 typedef long long ll; 6 typedef pair<int, int> pii; 7 8 char s[200010]; 9 int a[200010]; 10 int main() { 11 int m; 12 scanf("%s%d", s + 1, &m); 13 int n = strlen(s + 1); 14 for (int i = 1; i <= n; i++) 15 s[i + n] = s[i]; 16 for (int i = 1; i <= 2 * n; i++) 17 if (s[i] == 'E') 18 a[++a[0]] = i; 19 if (!a[0]) { 20 printf("0\n"); 21 return 0; 22 } 23 int last = 1; 24 ll ans = 0; 25 for (int i = 1; i <= n; i++) { 26 if (a[last] < i && last <= a[0]) 27 last++; 28 if (last > a[0]) 29 break; 30 ans += max(0, i + m - a[last]); 31 } 32 printf("%lld\n", ans); 33 }
F：
solver：lzh、zyh
1 #include <bits/stdc++.h> 2 using namespace std; 3 #define ff first 4 #define ss second 5 typedef long long ll; 6 typedef pair<int, int> pii; 7 8 ll a[5010][5010]; 9 const int mod = 1e9 + 7; 10 int main() { 11 int n, m; 12 scanf("%d%d", &n, &m); 13 if (n == 1) 14 printf("1\n"); 15 else if (n == 2) 16 printf("%d\n", m - 1); 17 else { 18 for (int i = 1; i <= 5000; i++) 19 a[1][i] = 1, a[2][i] = i - 1; 20 for (int i = 3; i <= n; i++) 21 for (int j = i; j <= m; j++) { 22 int tmp = j - 1; 23 while (tmp > 0) { 24 a[i][j] = (a[i][j] + a[i - 1][tmp] * 2 % mod - a[i - 2][tmp - 1] + mod) % mod; 25 tmp -= i; 26 } 27 } 28 printf("%lld\n", a[n][m]); 29 } 30 }
G：
solver：zyh、czq
1 #include <iostream> 2 #include <cstring> 3 #include <algorithm> 4 #include <vector> 5 using namespace std; 6 #define STRMAXLEN 500000 7 struct SAM { 8 struct node { 9 int parent, next[26], len; 10 int cnt; 11 }; 12 node tr[STRMAXLEN]; 13 int sz[STRMAXLEN]; 14 int rnk[STRMAXLEN]; 15 int last = 0; 16 int num = 0; 17 void init() { 18 last = 0; 19 num = 0; 20 tr[0].parent = -1; 21 } 22 void add(char ch) { 23 int p = last; 24 int t = ch - 'A'; 25 int now = ++num; 26 tr[now].len = tr[last].len + 1; 27 tr[now].cnt = 1; 28 while (p != -1 && tr[p].next[t] == 0) { 29 tr[p].next[t] = now; 30 p = tr[p].parent; 31 } 32 if (p != -1) { 33 int q = tr[p].next[t]; 34 if (tr[p].len + 1 == tr[q].len) { 35 tr[num].parent = q; 36 } else { 37 tr[++num].parent = tr[q].parent; 38 memcpy(tr[num].next, tr[q].next, sizeof(tr[num].next)); 39 tr[num].len = tr[p].len + 1; 40 while (p != -1 && tr[p].next[t] == q) { 41 tr[p].next[t] = num; 42 p = tr[p].parent; 43 } 44 tr[now].parent = tr[q].parent = num; 45 } 46 } else tr[now].parent = 0; 47 last = now; 48 } 49 void solve() { 50 //依赖排序，可以按这个顺序dp 51 for (int i = 0; i <= num; ++i) sz[i] = 0; 52 for (int i = 0; i <= num; ++i) sz[tr[i].len]++; 53 for (int i = 1; i <= num; ++i) sz[i] += sz[i - 1]; 54 for (int i = 0; i <= num; ++i) rnk[--sz[tr[i].len]] = i; 55 for (int i = num; i >= 1; --i) { 56 tr[tr[rnk[i]].parent].cnt += tr[rnk[i]].cnt; 57 } 58 } 59 int getAns(char ch[], int len) { 60 int ans = 1; 61 int now = 0; 62 for (int i = 0; i < len; ++i) { 63 now = tr[now].next[ch[i] - 'A']; 64 if (now == 0) { 65 ++ans; 66 now = tr[now].next[ch[i] - 'A']; 67 if (now == 0) return -1; 68 } 69 //cout<<now<<' '; 70 } 71 //cout<<endl; 72 return ans; 73 } 74 }; 75 SAM sam; 76 char str[500001]; 77 int main() { 78 scanf("%s", str); 79 sam.init(); 80 int len = strlen(str); 81 for (int i = 0; i < len; ++i) sam.add(str[i]); 82 int n; 83 scanf("%d", &n); 84 while (n--) { 85 scanf("%s", str); 86 len = strlen(str); 87 printf("%d\n", sam.getAns(str, len)); 88 } 89 }
I：
solver：czq
1 /* basic header */ 2 #include <bits/stdc++.h> 3 /* define */ 4 #define ll long long 5 #define pb emplace_back 6 #define mp make_pair 7 #define eps 1e-8 8 #define lson (curpos<<1) 9 #define rson (curpos<<1|1) 10 /* namespace */ 11 using namespace std; 12 /* header end */ 13 14 const int maxn = 2020; 15 const ll mod = 1e9 + 7; 16 vector<ll>edge[maxn]; 17 ll n, m, dp[maxn], cnt = 0, ans = 0; 18 19 ll dfs(ll cur) { 20 if (dp[cur] != -1) return dp[cur]; 21 if (cur > m) { 22 cnt++; 23 dp[cur] = 1; 24 return dp[cur]; 25 } 26 ll sum = 0; 27 for (auto x : edge[cur]) sum = ((dfs(x) % mod) + sum) % mod; 28 dp[cur] = sum; 29 return sum; 30 } 31 32 int main() { 33 for (int i = 0; i < maxn; i++) dp[i] = -1; 34 scanf("%lld%lld", &n, &m); 35 for (int i = 1; i <= m; i++) { 36 ll x; scanf("%lld", &x); 37 while (x--) { 38 ll y; scanf("%lld", &y); 39 edge[i].pb(y); 40 } 41 sort(edge[i].begin(), edge[i].end()); 42 } 43 ll ans = dfs(1); 44 printf("%lld %lld\n", ans, cnt); 45 return 0; 46 }
K：
solver：lzh、zyh、czq
1 #include <bits/stdc++.h> 2 using namespace std; 3 #define ff first 4 #define ss second 5 typedef long long ll; 6 typedef pair<int, int> pii; 7 8 char s[10010]; 9 int main() { 10 scanf("%s", s + 1); 11 int n = strlen(s + 1); 12 vector<ll> ans, noteql; 13 for (int i = 1; i < n; i++) 14 if (s[i] != s[i + 1]) 15 noteql.push_back(2 * i + 1); 16 printf("%d\n", noteql.size()); 17 if (noteql.size() == 0) { 18 if (s[1] == 'A') 19 printf("-1\n"); 20 else 21 printf("1\n"); 22 return 0; 23 } 24 ans.push_back(1), ans.push_back(-noteql.back()), noteql.pop_back(); 25 26 for (auto i : noteql) { 27 vector<ll> tmp = ans; 28 tmp.push_back(0); 29 for (int j = 1; j < tmp.size(); j++) 30 tmp[j] -= i * ans[j - 1]; 31 ans = tmp; 32 } 33 ll f = 1; 34 if (((noteql.size() + 1) % 2 == 0 && s[1] == 'A') || ((noteql.size() + 1) % 2 == 1 && s[1] == 'H')) 35 f = -1; 36 // for (auto i : ans) 37 // printf("%d ", f * i); 38 for (int i = 0; i < (int)ans.size(); i++) { 39 printf("%lld%c", f * ans[i], i == (int)ans.size() - 1 ? '\n' : ' '); 40 } 41 return 0; 42 }
L：
solver：lzh、zyh
1 #include <bits/stdc++.h> 2 using namespace std; 3 #define ff first 4 #define ss second 5 typedef long long ll; 6 typedef pair<int, int> pii; 7 8 char s[1010][1010]; 9 int a[1010][1010], b[1010]; 10 int d[20][10000]; 11 int query(int l, int r) { 12 int t = log2(r - l + 1); 13 return min(d[t][l], d[t][r - (1 << t) + 1]); 14 } 15 void init(int len) { 16 for (int i = 0; i < len; ++i) { 17 d[0][i] = b[i]; 18 } 19 int t = 1; 20 for (int i = 1; t <= len; ++i) { 21 for (int j = 0; j + t < len; ++j) 22 d[i][j] = min(d[i - 1][j], d[i - 1][j + t]); 23 t <<= 1; 24 } 25 } 26 int main() { 27 int n, m, ans = 0; 28 scanf("%d%d", &n, &m); 29 for (int i = 1; i <= n; i++) { 30 scanf("%s", s[i] + 1); 31 for (int j = 1; j <= m; j++) { 32 if (s[i][j] == 'G') 33 a[i][j]++; 34 a[i][j] += a[i][j - 1]; 35 } 36 } 37 for (int j = 1; j <= m; j++) { 38 for (int i = 1; i <= n; i++) { 39 int l = j, r = m, mid, tmp = 0; 40 while (l <= r) { 41 mid = l + r >> 1; 42 if (a[i][mid] - a[i][j - 1] == 0 || a[i][mid] - a[i][j - 1] == mid - j + 1) 43 l = mid + 1, tmp = max(tmp, mid); 44 else 45 r = mid - 1; 46 } 47 b[i - 1] = tmp - j + 1; 48 } 49 init(n); 50 for (int i = 1; i <= n; i++) { 51 int l = 1, r = n - i + 1, mid, tmp = 1; 52 while (l <= r) { 53 mid = l + r >> 1; 54 if (query(i - 1, i + mid - 2) >= mid) 55 l = mid + 1, tmp = max(tmp, mid); 56 else 57 r = mid - 1; 58 } 59 ans = max(ans, tmp); 60 } 61 } 62 printf("%d\n", ans * ans); 63 }
M：
solver：lzh
1 #include <bits/stdc++.h> 2 using namespace std; 3 #define ff first 4 #define ss second 5 typedef long long ll; 6 typedef pair<int, int> pii; 7 8 int a[1010]; 9 int main() { 10 int n, x; 11 cin >> n >> x; 12 for (int i = 1; i <= n; i++) 13 cin >> a[i]; 14 int ans = 1; 15 for (int i = 1; i <= n; i++) { 16 int tmp = 1; 17 for (int j = i + 1; j <= n; j++) 18 if (a[j] - a[j - 1] <= x) 19 tmp++; 20 else 21 break; 22 ans = max(ans, tmp); 23 } 24 cout << ans << endl; 25 }