Codeforces 999
A
/*Huyyt*/ #include<bits/stdc++.h> #define mem(a,b) memset(a,b,sizeof(a)) using namespace std; typedef long long ll; typedef unsigned long long ull; const int dir[8][2] = {{0, 1}, {1, 0}, {0, -1}, { -1, 0}, {1, 1}, {1, -1}, { -1, -1}, { -1, 1}}; const int mod = 1e9 + 7, gakki = 5 + 2 + 1 + 19880611 + 1e9; const int MAXN = 2e5 + 5, MAXM = 2e5 + 5, N = 2e5 + 5; const int MAXQ = 100010; /*int to[MAXM << 1], nxt[MAXM << 1], Head[MAXN], tot = 1; inline void addedge(int u, int v) { to[++tot] = v; nxt[tot] = Head[u]; Head[u] = tot; }*/ inline void read(int &v) { v = 0; char c = 0; int p = 1; while (c < '0' || c > '9') { if (c == '-') { p = -1; } c = getchar(); } while (c >= '0' && c <= '9') { v = (v << 3) + (v << 1) + c - '0'; c = getchar(); } v *= p; } int num[105]; int visit[105]; int main() { int n, k; read(n), read(k); for (int i = 1; i <= n; i++) { read(num[i]); } int anser = 0; for (int i = 1; i <= n; i++) { if (num[i] <= k) { visit[i] = 1; anser++; } else { break; } } for (int i = n; i >= 1; i--) { if (visit[i]) { break; } if (num[i] <= k) { anser++; } else { break; } } cout << anser << endl; return 0; }
B
/*Huyyt*/ #include<bits/stdc++.h> #define mem(a,b) memset(a,b,sizeof(a)) using namespace std; typedef long long ll; typedef unsigned long long ull; const int dir[8][2] = {{0, 1}, {1, 0}, {0, -1}, { -1, 0}, {1, 1}, {1, -1}, { -1, -1}, { -1, 1}}; const int mod = 1e9 + 7, gakki = 5 + 2 + 1 + 19880611 + 1e9; const int MAXN = 2e5 + 5, MAXM = 2e5 + 5, N = 2e5 + 5; const int MAXQ = 100010; /*int to[MAXM << 1], nxt[MAXM << 1], Head[MAXN], tot = 1; inline void addedge(int u, int v) { to[++tot] = v; nxt[tot] = Head[u]; Head[u] = tot; }*/ inline void read(int &v) { v = 0; char c = 0; int p = 1; while (c < '0' || c > '9') { if (c == '-') { p = -1; } c = getchar(); } while (c >= '0' && c <= '9') { v = (v << 3) + (v << 1) + c - '0'; c = getchar(); } v *= p; } string s; void doit(int x) { for (int i = 0; i <= x / 2 - 1; i++) { swap(s[i], s[x - 1 - i]); } } int main() { int n; read(n); cin >> s; for (int i = 2; i <= n; i++) { if (n % i == 0) { doit(i); } } cout << s << endl; return 0; }
C
/*Huyyt*/ #include<bits/stdc++.h> #define mem(a,b) memset(a,b,sizeof(a)) using namespace std; typedef long long ll; typedef unsigned long long ull; const int dir[8][2] = {{0, 1}, {1, 0}, {0, -1}, { -1, 0}, {1, 1}, {1, -1}, { -1, -1}, { -1, 1}}; const int mod = 1e9 + 7, gakki = 5 + 2 + 1 + 19880611 + 1e9; const int MAXN = 2e5 + 5, MAXM = 2e5 + 5, N = 2e5 + 5; const int MAXQ = 100010; /*int to[MAXM << 1], nxt[MAXM << 1], Head[MAXN], tot = 1; inline void addedge(int u, int v) { to[++tot] = v; nxt[tot] = Head[u]; Head[u] = tot; }*/ inline void read(int &v) { v = 0; char c = 0; int p = 1; while (c < '0' || c > '9') { if (c == '-') { p = -1; } c = getchar(); } while (c >= '0' && c <= '9') { v = (v << 3) + (v << 1) + c - '0'; c = getchar(); } v *= p; } string s; int visit[400005]; queue<int> ch[30]; int main() { int n, k; read(n), read(k); cin >> s; for (int i = 0; i < s.size(); i++) { ch[s[i] - 'a'].push(i); //cout<<s[i]-'a'<<" "<<i<<endl; } for (int j = 1; j <= k; j++) { for (int i = 0; i <= 25; i++) { if (ch[i].size() > 0) { visit[ch[i].front()] = 1; ch[i].pop(); break; } } } for (int i = 0; i < s.size(); i++) { if(!visit[i]) { cout<<s[i]; } } return 0; }
D
磨了半天 没想到直接暴力也能过
#include <bits/stdc++.h> using namespace std; typedef long long LL; const int N = 202020; int n, m, a[N]; vector<int> c[N]; LL ans; int main() { scanf("%d%d", &n, &m); for (int i = 0; i < n; i++) { scanf("%d", &a[i]); c[a[i] % m].push_back(i); } int ptr = 0; int goal = n / m; for (int i = 0; i < m; i++) //从0开始检查比goal大的 { while ((int)c[i].size() > goal) //如果枚举到的比水平线高 { ptr = max(ptr, i); //保证找到的要在其后 while ((int)c[ptr % m].size() >= goal) //找到小于水平线的 { ptr ++; } int tk = min((int)c[i].size() - goal, goal - (int)c[ptr % m].size()); //取多出与缺少的最小值 int dlt = ptr - i; //多的和少的之间的距离 while (tk--) { ans += dlt; a[c[i].back()] += dlt; c[ptr % m].push_back(c[i].back()); c[i].pop_back(); } } } printf("%lld\n", ans); for (int i = 0; i < n; i++) { printf("%d%c", a[i], " \n"[i + 1 == n]); } }
E
暴力染色即可
/*Huyyt*/ #include<bits/stdc++.h> #define mem(a,b) memset(a,b,sizeof(a)) using namespace std; typedef long long ll; typedef unsigned long long ull; const int dir[8][2] = {{0, 1}, {1, 0}, {0, -1}, { -1, 0}, {1, 1}, {1, -1}, { -1, -1}, { -1, 1}}; const int mod = 1e9 + 7, gakki = 5 + 2 + 1 + 19880611 + 1e9; const int MAXN = 5e3 + 5, MAXM = 5e3 + 5, N = 2e5 + 5; const int MAXQ = 100010; int to[MAXM << 1], nxt[MAXM << 1], Head[MAXN], tot = 1; inline void addedge(int u, int v) { to[++tot] = v; nxt[tot] = Head[u]; Head[u] = tot; } inline void read(int &v) { v = 0; char c = 0; int p = 1; while (c < '0' || c > '9') { if (c == '-') { p = -1; } c = getchar(); } while (c >= '0' && c <= '9') { v = (v << 3) + (v << 1) + c - '0'; c = getchar(); } v *= p; } int du[MAXN], visit[MAXN]; int visit2[MAXN]; int ans = 0; void dfs(int x, int pre) { visit[x] = pre; visit2[x] = 1; for (int i = Head[x]; i; i = nxt[i]) { int v = to[i]; if (!visit2[v]) { dfs(v, pre); } } } int vis[MAXN]; int main() { int n, m, s, u, v; read(n), read(m), read(s); for (int i = 1; i <= m; i++) { read(u), read(v); addedge(u, v); du[v]++; } for (int i = 1; i <= n; i++) { if (!visit[i]) { mem(visit2, 0); dfs(i, i); } } mem(visit2,0); dfs(s, s); for (int i = 1; i <= n; i++) { if (vis[visit[i]] == 0) { vis[visit[i]] = 1; ans++; } } cout << ans - 1 << endl; }
tarjan缩点后找入度为0的点 可以O(n+m)的做
F
dp
#include <bits/stdc++.h> using namespace std; typedef long long LL; const int N = 101010; int n, k, c[N], f[N], h[N]; const int K = 514; int dp[10 * K]; int cal(int x) { if (f[x] == 0) { return 0; } int atmst = min(f[x] * k, c[x]); for (int i = 0; i <= atmst; i++) { dp[i] = 0; } while (f[x]--) { for (int i = atmst; i >= 0; i--) for (int j = 1; j <= k and i - j >= 0; j++) { dp[i] = max(dp[i], dp[i - j] + h[j]); } } return dp[atmst]; } int main() { scanf("%d%d", &n, &k); for (int i = 0; i < n * k; i++) { int ci; scanf("%d", &ci); c[ci]++; } for (int i = 0; i < n; i++) { int fi; scanf("%d", &fi); f[fi]++; } for (int i = 1; i <= k; i++) { scanf("%d", &h[i]); } int ans = 0; for (int i = 0; i < N; i++) { ans += cal(i); } cout << ans << endl; }
