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 }
来源：https://www.cnblogs.com/JHSeng/p/12347673.html