Codeforces Round #267 (Div. 2)
Codeforces Round #267 (Div. 2)
A:签到题。直接for一遍
B:取异或就是不同的数。然后bitcount一下推断就可以
C:dp,dp[i]表示到i的最大值,然后对取与不取当前位置进行转移就可以,要先把前缀和预处理出来
D:先利用map。把字符串hash掉。然后建图,现场在做的时候是直接记忆化搜索,只是这样处理不了环的情况,果断fst了,后来换了下姿势。先求强连通进行缩点,缩点完就是一个dag了。然后记忆化搜索就可以
E:贪心+单调性维护,利用单调栈维护,每次遇到一个同样的数字的时候,中间一段就弹出栈,记录下匹配数字,然后下次遇到这个数字的时候,就能够优先组成一个,然后从之后的位置開始继续找
代码:
A:
#include <cstdio> int main() { int n, p, q; int ans = 0; scanf("%d", &n); while (n--) { scanf("%d%d", &p, &q); if (q - p >= 2) ans++; } printf("%d\n", ans); return 0; }
B:
#include <cstdio> #include <cstring> int n, m, k; int bitcount(int x) { int ans = 0; while (x) { ans += (x&1); x >>= 1; } return ans; } const int N = 1005; int x[N]; int main() { scanf("%d%d%d", &n, &m, &k); for (int i = 0; i <= m; i++) scanf("%d", &x[i]); int ans = 0; for (int i = 0; i < m; i++) { if (bitcount(x[i]^x[m]) <= k) ans++; } printf("%d\n", ans); return 0; }
C:
#include <cstdio> #include <cstring> #include <algorithm> using namespace std; const int N = 5005; const long long INF = 0x3f3f3f3f3f3f3f; long long dp[N][N]; int n, m, k; long long num[N], pre[N]; int main() { scanf("%d%d%d", &n, &m, &k); for (int i = 1; i <= n; i++) { scanf("%lld", &num[i]); pre[i] = pre[i - 1] + num[i]; } for (int i = 0; i <= n; i++) for (int j = 0; j <= k; j++) dp[i][j] = -INF; dp[0][0] = 0; for (int i = 1; i <= n; i++) { for (int j = 0; j <= k; j++) { if (j && i >= m) dp[i][j] = max(dp[i][j], dp[i - m][j - 1] + pre[i] - pre[i - m]); dp[i][j] = max(dp[i][j], dp[i - 1][j]); } } printf("%lld\n", dp[n][k]); return 0; }
D:
#include <cstdio> #include <cstring> #include <vector> #include <stack> #include <algorithm> #include <string> #include <iostream> #include <map> using namespace std; typedef long long ll; const int N = 1000005; vector<int> g[N], scc[N]; int pre[N], lowlink[N], sccno[N], dfs_clock, scc_cnt; stack<int> S; typedef pair<ll, ll> pii; int n, m, hn; map<string, int> hash; string str[N]; pii s[N]; pii p[N]; pii dp[N]; int vis[N]; vector<int> g2[N]; void dfs_scc(int u) { pre[u] = lowlink[u] = ++dfs_clock; S.push(u); for (int i = 0; i < g[u].size(); i++) { int v = g[u][i]; if (!pre[v]) { dfs_scc(v); lowlink[u] = min(lowlink[u], lowlink[v]); } else if (!sccno[v]) lowlink[u] = min(lowlink[u], pre[v]); } if (lowlink[u] == pre[u]) { scc_cnt++; pii tmp = s[S.top()]; while (1) { int x = S.top(); S.pop(); tmp = min(tmp, s[x]); sccno[x] = scc_cnt; if (x == u) break; } p[scc_cnt] = tmp; } } void find_scc(int n) { dfs_clock = scc_cnt = 0; memset(sccno, 0, sizeof(sccno)); memset(pre, 0, sizeof(pre)); for (int i = 0; i < n; i++) if (!pre[i]) dfs_scc(i); for (int i = 0; i < n; i++) { for (int j = 0; j < g[i].size(); j++) { int u = sccno[i], v = sccno[g[i][j]]; if (u == v) continue; g2[u].push_back(v); } } } int get(string& str) { for (int i = 0; i < str.length(); i++) if (str[i] >= 'A' && str[i] <= 'Z') str[i] = str[i] - 'A' + 'a'; if (!hash.count(str)) { hash[str] = hn; int cnt = 0; for (int i = 0; i < str.length(); i++) if (str[i] == 'r') cnt++; s[hn].first = cnt; s[hn].second = str.length(); hn++; } return hash[str]; } pii dfs(int u) { if (vis[u]) return dp[u]; vis[u] = 1; dp[u] = p[u]; for (int i = 0; i < g2[u].size(); i++) { int v = g2[u][i]; dp[u] = min(dp[u], dfs(v)); } return dp[u]; } int main() { scanf("%d", &n); for (int i = 0; i < n; i++) { cin >> str[i]; get(str[i]); } scanf("%d", &m); string u, v; while (m--) { cin >> u >> v; int uu = get(u); int vv = get(v); g[uu].push_back(vv); } find_scc(hn); ll ans1 = 0, ans2 = 0; for (int i = 0; i < n; i++) { int u = sccno[get(str[i])]; dfs(u); ans1 += dp[u].first; ans2 += dp[u].second; } cout << ans1 << " " << ans2 << endl; return 0; }
E:
#include <cstdio> #include <cstring> #include <map> #include <vector> #include <algorithm> #include <stack> using namespace std; const int N = 500005; int n, b[N]; void solve() { vector<int> ans; int u = 0; for (; u < n; u++) { map<int, int> pre, sum; stack<int> s; for (int &i = u; i < n; i++) { if (pre[b[i]]) { for (int j = 0; j < 2; j++) { ans.push_back(pre[b[i]]); ans.push_back(b[i]); } break; } while (!s.empty() && (sum[b[i]] > 1 || sum[b[i]] == 1 && b[i] != s.top())) { pre[s.top()] = b[i]; sum[s.top()]--; s.pop(); } s.push(b[i]); sum[b[i]]++; } } printf("%d\n", ans.size()); for (int i = 0; i < ans.size(); i++) printf("%d%c", ans[i], ans.size() - 1 == i? '\n' : ' '); } int main() { scanf("%d", &n); for (int i = 0; i < n; i++) scanf("%d", &b[i]); solve(); return 0; }