Codeforces Round #467 (Div. 2)

输出除0以外的数字种数即可。

猜想答案离y不会很远。暴力枚举答案， $O(\sqrt n)$验证，如果有因数落在区间$[2,p]$里就不合法。

二分答案搞一搞。

如果走到一个出度为零而且该对面走的点p就赢了。如果有环就可以平局。否则就输了。

找到点p之后往回走走到起始点就行了，记录下这条路径。

先坑着。

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<set>
 4 using namespace std;
 5 int main() {
 6     set < int > s;
 7     int n; scanf("%d", &n);
 8     for (int i = 0, t; i < n; ++i) {
 9         scanf("%d", &t);
10         if (t) s.insert(t);
11     }
12     cout << s.size();
13     return 0;
14 }

A

 1 #include<cstdio>
 2 #include<iostream>
 3 using namespace std;
 4 int p, y;
 5 bool check(int x) {
 6     for (int i = 1; i * i <= x; ++i) if (x % i == 0) {
 7         if (i >= 2 && i <= p) return false;
 8         if (x / i >= 2 && x / i <= p) return false;
 9     }
10     return true;
11 }
12 int main() {
13     cin >> p >> y;
14     int ans = -1;
15     for (int i = y; i >= max(2, y - 2000); --i) {
16         if (check(i)) {
17             ans = i; break;
18         }
19     }
20     printf("%d\n", ans);
21     return 0;
22 }

B

 1 #include<cstdio>
 2 #include<cmath>
 3 #include<iostream>
 4 using namespace std;
 5 typedef double db;
 6 typedef long long ll;
 7 ll k, d, t, T, x1, x2;
 8 bool check(db x) {
 9     ll p = (ll)x / T;
10     db a = x1 * p, b = x2 * p;
11     x -= p * T;
12     a += min(x, (db)x1);
13     x -= x1;
14     b += max(.0, x);
15     return a * 2 + b >= t * 2;
16 }
17 int main() {
18     cin >> k >> d >> t;
19     T = (k + d - 1) / d * d;
20     x1 = (k / d) * d + (k % d);
21     x2 = T - x1;
22     double l = 0, r = t * 2;
23     for (int i = 0; i < 1000; ++i) {
24         double m = (l + r) / 2;
25         if (check(m)) r = m;
26         else l = m;
27     }
28     printf("%.15lf\n", l);
29     return 0;
30 }

C

 1 #include<queue>
 2 #include<cstdio>
 3 #include<vector>
 4 #include<iostream>
 5 #define pb push_back
 6 #define mp make_pair
 7 using namespace std;
 8 const int N = 100005;
 9 int n, m, vis[N][2], out[N], vs[N], son[N][2];
10 vector < int > g[N], r[N];
11 void dfs(int u, int d) {
12     vis[u][d] = 1;
13     for (int v, i = 0; i < g[u].size(); ++i) {
14         v = g[u][i];
15         if (!vis[v][!d]) dfs(v, !d);
16     }
17 }
18 bool dfs(int u) {
19     if (vs[u]) return true;
20     vs[u] = 1;
21     for (int i = 0; i < g[u].size(); ++i)
22         if (dfs(g[u][i])) return true;
23     vs[u] = 0;
24     return false;
25 }
26 bool circle() {
27     return dfs(m);
28 }
29 int main() {
30     scanf("%d%d", &n, &m);
31     for (int c, i = 1; i <= n; ++i) {
32         scanf("%d", &c); out[i] = c;
33         for (int t; c--;) {
34             scanf("%d", &t);
35             g[i].pb(t);
36             r[t].pb(i);
37         }
38     }
39     scanf("%d", &m);
40     dfs(m, 0);
41     int p = 0;
42     for (int i = 1; i <= n; ++i) if (!out[i]) {
43         if (vis[i][1]) {p = i; break;}
44     }
45     if (p) {
46         queue < pair < int , int > > q;
47         q.push(mp(p, 1));
48         while (!q.empty()) {
49             int u = q.front().first, s = q.front().second; q.pop();
50             for (int i = 0; i < r[u].size(); ++i) {
51                 int v = r[u][i];
52                 if (vis[v][!s] && !son[v][!s]) {
53                     son[v][!s] = u; q.push(mp(v, !s));
54                 }
55             }
56         }
57         puts("Win");
58         for (int u = m, s = 1; u; s ^= 1, u = son[u][s]) {
59             printf("%d ", u);
60         } printf("\n");
61     } else {
62         if (circle()) puts("Draw");
63         else puts("Lose");
64     }
65     return 0;
66 }

D

原文地址：https://www.cnblogs.com/p0ny/p/8472467.html