比赛链接

Dashboard - Codeforces Round 871 (Div. 4) - Codeforces

A. Love Story

找到与codeforces 有多少个不同的字符。

#include<bits/stdc++.h>
#define int long long  
#define TEST int T; cin >> T; while (T--)
#define ios ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr)
#define pll pair<int,int>
const int N = 1e6 + 30;
const int M = 1e3 + 10;
const int inf = 512785182741247112;
const int mod = 1e9 + 7;
using namespace std;
string s = "codeforces";
void solve() {string st;cin >> st;int ans = 0;for (int i = 0; i < st.size(); i++){if (st[i] != s[i])ans++;}cout << ans << "\n";}
signed main() {ios;TESTsolve();return 0;
}

B. Blank Space

找连续最长的0.

#include<bits/stdc++.h>
#define int long long  
#define TEST int T; cin >> T; while (T--)
#define ios ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr)
#define pll pair<int,int>
const int N = 1e6 + 30;
const int M = 1e3 + 10;
const int inf = 512785182741247112;
const int mod = 1e9 + 7;
using namespace std;void solve() {int n;cin >> n;int cnt = 0, ans = 0;for (int i = 1; i <= n; i++){int x;cin >> x;if (x == 0) cnt++;else{ans = max(ans, cnt);cnt = 0;}}ans = max(ans, cnt);cout << ans << "\n";
}
signed main() {ios;TESTsolve();return 0;
}

C. Mr. Perfectly Fine

分成01、10、11的三种情况，找全部的最小值，前面两个的最小值相加与第三个的最小值比较出最小值。

#include<bits/stdc++.h>
#define int long long  
#define TEST int T; cin >> T; while (T--)
#define ios ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr)
#define pll pair<int,int>
const int N = 1e6 + 30;
const int M = 1e3 + 10;
const int inf = 512785182741247112;
const int mod = 1e9 + 7;
using namespace std;void solve() {int n;cin >> n;int c1 = 1e9, c2 = 1e9, c3 = 1e9;for (int i = 1; i <= n; i++){int t;cin >> t;string s;cin >> s;if (s == "01") c2 = min(c2, t);if (s == "10") c1 = min(c1, t);if (s == "11") c3 = min(c3, t);}int res = min(c1 + c2, c3);if (res != 1e9)cout << min(c1 + c2, c3) << "\n";else cout << "-1\n";
}
signed main() {ios;TESTsolve();return 0;
}

D. Gold Rush

按照题意可以将一堆分成三堆，看看目标是否出现即可，搜索。

#include<bits/stdc++.h>
#define int long long  
#define TEST int T; cin >> T; while (T--)
#define ios ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr)
#define pll pair<int,int>
const int N = 1e6 + 30;
const int M = 1e3 + 10;
const int inf = 512785182741247112;
const int mod = 1e9 + 7;
using namespace std;void solve() {int n, m;cin >> n >> m;if (m > n){cout << "NO\n";return;}if (n == m){cout << "YES\n";return;}queue<int>q;q.push(n);while (q.size()){int k = q.front();q.pop();if (k == m){cout << "YES\n";return;}if (k < m|| k % 3 != 0) continue;int t=k / 3;q.push(t);q.push(t * 2);}cout << "NO\n";
}signed main() {ios;TESTsolve();return 0;
}

E. The Lakes

dfs染色法。

#include<bits/stdc++.h>
#define int long long  
#define TEST int T; cin >> T; while (T--)
#define ios ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr)
#define pll pair<int,int>
const int N = 1e6 + 30;
const int M = 1e3 + 10;
const int inf = 512785182741247112;
const int mod = 1e9 + 7;
using namespace std;
int vis[1001][1001];
int a[1001][1001];
int q[N];
int dx[] = { 1,0,-1,0 };
int dy[] = { 0,1,0,-1 };
int n, m;
int sum = 0;
int ans = 0;
void dfs(int x, int y, int tag)
{vis[x][y] = tag;sum += a[x][y];ans = max(ans, sum);for (int i = 0; i < 4; i++){int tx = x + dx[i];int ty = y + dy[i];if (tx<1 || ty<1 || tx>n || ty>m) continue;if (a[tx][ty] == 0) continue;if (vis[tx][ty]) continue;dfs(tx, ty, tag);}
}
void solve() {cin >> n >> m;ans = 0;for (int i = 1; i <= n; i++){for (int j = 1; j <= m; j++){cin >> a[i][j];}}int cnt = 0;for (int i = 1; i <= n; i++){for (int j = 1; j <= m; j++){if (a[i][j] == 0) continue;if (vis[i][j]) continue;sum = 0;++cnt;dfs(i, j, cnt);}}cout << ans << "\n";for (int i = 1; i <= n; i++){for (int j = 1; j <= m; j++){vis[i][j] = 0;}}
}signed main() {ios;TESTsolve();return 0;
}

F. Forever Winter

模拟一下，与出入度为1的点连接的就是与主节点连接的点，输出他们的出入度即可，与主节点连接的点，出入度需要减一。

#include<bits/stdc++.h>
#define int long long  
#define TEST int T; cin >> T; while (T--)
#define ios ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr)
#define pll pair<int,int>
const int N = 1e6 + 30;
const int M = 1e3 + 10;
const int inf = 512785182741247112;
const int mod = 1e9 + 7;
using namespace std;
void solve() {int n, m;cin >> n >> m;vector<vector<int>>f(n + 1);for (int i = 1; i <= m; i++){int u, v;cin >> u >> v;f[u].push_back(v);f[v].push_back(u);}set<int>q;for (int i = 1; i <= n; i++){if (f[i].size() == 1){q.insert(f[i][0]);}}int ans1;for (auto x : q){ans1 = f[x].size() - 1;break;}int ans2;for (int i = 1; i <= n; i++){set<int>tt;for (auto x : f[i]) tt.insert(x);if (tt == q){ans2 = f[i].size();break;}}cout << ans2 << ' ' << ans1 << "\n";
}signed main() {ios;TESTsolve();return 0;
}

G. Hits Different

一开始用搜索，优化半天还是超时，正解是二位数组前缀和dp。

#include<bits/stdc++.h>
#define int long long  
#define TEST int T; cin >> T; while (T--)
#define ios ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr)
#define pll pair<int,int>
const int N = 1e6 + 30;
const int M = 1e3 + 10;
const int inf = 512785182741247112;
const int mod = 1e9 + 7;
using namespace std;
int a[1500][1500];
int cnt=1;
int ans[N];
void solve() {int n;cin >> n;cout << ans[n] << "\n";
}signed main() {for (int i = 1; i < 1500; i++){for (int j = i - 1; j >= 1; j--){a[j][i - j] = a[j - 1][i - j] + a[j][i - j - 1] - a[j - 1][i - j - 1] + cnt * cnt;ans[cnt] = a[j][i - j];cnt++;}}ios;TESTsolve();return 0;
}

H. Don't Blame Me

找子序列所有元素的与和的二进制里1的个数等于k，数位dp。

#include<bits/stdc++.h>
#define int long long  
#define TEST int T; cin >> T; while (T--)
#define ios ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr)
#define pll pair<int,int>
const int N = 1e6 + 30;
const int M = 1e3 + 10;
const int inf = 512785182741247112;
const int mod = 1e9 + 7;
using namespace std;
int dp[N][64];
int a[N];
void solve()
{int n, k;cin >> n >> k;for (int i = 1; i <= n; i++){cin >> a[i];}for (int i = 1; i <= n; i++){dp[i][a[i]] = 1;for (int j = 0; j <= 63; j++){dp[i][j] = (dp[i][j] + dp[i - 1][j]) % mod;dp[i][j & a[i]] = (dp[i][j & a[i]] + dp[i - 1][j]) % mod;}}int ans = 0;for (int i = 0; i <= 63; i++){int cnt = 0;for (int j = 0; j < 6; j++){if ((i >> j) & 1){cnt++;}}if (cnt == k){ans = (ans + dp[n][i]) % mod;}}cout << ans << "\n";for (int i = 1; i <= n; i++){for (int j = 0; j <= 63; j++){dp[i][j] = 0;}}
}
signed main() {ios;TESTsolve();return 0;
}