J题队友犯了初始化的错，白给6发，本来能1A的
B：
solver：lzh、czq
就是个拓扑排序
1 #include <bits/stdc++.h>
2 using namespace std;
3 #define ff first
4 #define ss second
5 #define mp make_pair
6 typedef long long ll;
7 typedef pair<int, int> pii;
8
9 vector<int> g[400010];
10 vector<int> v;
11 int a[400010], deg[400010];
12 int main() {
13 v.push_back(0);
14 int n;
15 scanf("%d", &n);
16 for (int i = 0; i < n; i++) {
17 int j;
18 scanf("%d%d", &a[i], &j);
19 while (j--) {
20 int x;
21 scanf("%d", &x);
22 g[i].push_back(x - 1);
23 deg[x - 1]++;
24 }
25 }
26 int ans = 0;
27 priority_queue<pii> pq;
28 for (int i = 0; i < n; i++)
29 if (!deg[i]) pq.push(mp(-a[i], i));
30 while (!pq.empty()) {
31 pii x = pq.top();
32 pq.pop();
33 v.push_back(-x.ff);
34 for (auto i : g[x.ss]) {
35 deg[i]--;
36 if (!deg[i])
37 pq.push(mp(-a[i], i));
38 }
39 }
40 reverse(v.begin(), v.end());
41 for (int i = 0; i < n; i++) ans = max(ans, v[i] + i);
42 printf("%d\n", ans);
43 }
C：
solver：czq
直接把圆周切个1000份
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 = 1010;
15 const double pi = acos(-1);
16 vector<pair<double, double>>ans;
17 vector<int>g[maxn];
18 int n, __size[maxn];
19
20 void dfs(int u, int father, double angle) {
21 __size[u]++;
22 for (auto v : g[u]) {
23 if (v != father) {
24 ans[v].first = ans[u].first + cos(angle);
25 ans[v].second = ans[u].second + sin(angle);
26 dfs(v, u, angle);
27 __size[u] += __size[v];
28 angle += pi / maxn * __size[v];
29 }
30 }
31 }
32
33 int main() {
34 scanf("%d", &n);
35 ans.resize(n + 1);
36 for (int i = 1; i < n; i++) {
37 int u, v; scanf("%d%d", &u, &v);
38 g[u].pb(v), g[v].pb(u);
39 }
40 dfs(1, -1, 0);
41 for (int i = 1; i <= n; i++) printf("%.10f %.10f\n", ans[i].first, ans[i].second);
42 return 0;
43 }
G：
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[30];
9 int notuse[30];
10 const int dd[4][2] = { -1, 0, 1, 0, 0, 1, 0, -1 };
11 const int C[4][2] = { 0, 1, 1, 0, 2, 3, 3, 2 };
12 const char fa[] = { "LRUD" };
13 int main() {
14 scanf("%s", s + 1);
15 int n = strlen(s + 1);
16 if (n == 1) {
17 if (s[1] == ‘L‘)
18 printf("1 0\n0\n");
19 else if (s[1] == ‘R‘)
20 printf("-1 0\n0\n");
21 else if (s[1] == ‘U‘)
22 printf("0 -1\n0\n");
23 else
24 printf("0 1\n0\n");
25 return 0;
26 }
27 for (int j = 0; j <= 4; j += 2) {
28 vector<int> tmp;
29 for (int i = 1; i <= n; i++)
30 if (s[i] == fa[j] || s[i] == fa[j + 1])
31 tmp.push_back(i);
32 for (int i = tmp.size() - 3; i >= 0; i--)
33 if (tmp[i] + 1 == tmp[i + 1] && tmp[i + 1] + 1 == tmp[i + 2])
34 notuse[tmp[i]]++;
35 }
36 for (int j = 0; j <= 4; j += 2) {
37 int tmp = 0;
38 for (int i = n; i >= 1; i--)
39 if (s[i] == fa[j] || s[i] == fa[j + 1])
40 tmp++;
41 else
42 break;
43 if (tmp >= 3) {
44 printf("impossible\n");
45 return 0;
46 }
47 }
48
49 vector<pii> ans;
50 pii cur = { 0, 0 };
51 int step = n + 1;
52 for (int i = n; i >= 1; i--)
53 if (!notuse[i]) {
54 for (int j = 0; j < 4; j++)
55 if (s[i] == fa[j]) {
56 if (i != n)
57 ans.push_back({ cur.ff + dd[j][0], cur.ss + dd[j][1] });
58
59 pii tmp = { (1 << step) *(-dd[j][0]), (1 << step) *(-dd[j][1]) };
60 if (tmp.ff)
61 cur.ff += tmp.ff;
62 else
63 cur.ss += tmp.ss;
64 }
65 step--;
66 }
67 printf("%d %d\n%d\n", cur.ff, cur.ss, ans.size());
68 for (auto i : ans)
69 printf("%d %d\n", i.ff, i.ss);
70 }
H：
solver：zyh、lzh
1 #include <bits/stdc++.h>
2 using namespace std;
3 bool z[1000001];
4 int main() {
5 int n, c, b;
6 scanf("%d%d%d", &n, &c, &b);
7 for (int i = 0; i < b; ++i) {
8 int t;
9 scanf("%d", &t);
10 z[t - 1] = true;
11 }
12 int now = 0;
13 if (c & 1) {
14 now = 1;
15 c++;
16 }
17 for (int i = 0; i < n; ++i) {
18 if (z[i] || !c) printf("0");
19 else {
20 printf("%d", now);
21 c--;
22 now = 1 - now;
23 }
24 }
25 }
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 = 2e5 + 10;
15 int n, a[maxn];
16 double ans = INT_MAX;
17
18 int main() {
19 scanf("%d", &n);
20 for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
21 sort(a + 1, a + 1 + n);
22 for (int i = 1; i <= n; i++) {
23 if (a[i] > i) return puts("-1"), 0;
24 ans = min(ans, double(a[i]) / (i));
25 }
26 printf("%.10f\n", ans);
27 return 0;
28 }
J：
solver：zyh、czq
1 #include <bits/stdc++.h>
2 using namespace std;
3 long long p[200001];
4 int main() {
5 int n;
6 scanf("%d", &n);
7 for (int i = 1; i <= n; ++i) scanf("%lld", &p[i]);
8 sort(p + 1, p + n + 1);
9 p[0] = -1;
10 if (p[n] == p[n - 1]) {
11 printf("%lld\n", p[n] - p[n - 2]); return 0;
12 }
13 long long sum = p[n] - p[n - 1];
14 long long Last = p[n - 1];
15 long long maxT = __LONG_LONG_MAX__;
16 long long T = 1;
17 long long ans = 0;
18 int tot = 0;
19 for (int i = n - 1; i >= 1; --i) {
20 ++tot;
21 if (p[i] != p[i - 1]) {
22 if (maxT >= Last - p[i]) {
23 ans += T * (Last - p[i]);
24 sum -= (T - 1) * (Last - p[i]);
25 T = int(log2(tot)) + 1;
26 if (T > 1) maxT = sum / (T - 1);
27 else maxT = sum + 1;
28 Last = p[i];
29 } else break;
30 }
31 }
32 if (T > 1) ans += T * maxT + sum % (T - 1);
33 else ans += T * maxT;
34 printf("%lld", ans);
35 return 0;
36 }
K：
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 int n, m;
15 string a, b, tmp, ans = "";
16
17 int main() {
18 ios::sync_with_stdio(false);
19 cin.tie(0);
20 cin >> n >> m >> a >> b;
21 tmp = a;
22 reverse(b.begin(), b.end());
23 reverse(tmp.begin(), tmp.end());
24 int p = 0;
25 for (int i = 0; i < m - n; i++) {
26 int x = b[i] - tmp[p]; if (x < 0) x += 26;
27 ans += (char)(x + ‘a‘);
28 tmp += ans[p++];
29 }
30 tmp += a;
31 reverse(ans.begin(), ans.end());
32 cout << ans << a << endl;
33 return 0;
34 }
原文地址：https://www.cnblogs.com/JHSeng/p/12336837.html
时间： 2024-10-05 08:28:27