勤劳努力的小蜜蜂

A. Minimize! 输出\(b-a\)即可 code: 123456789101112131415161718#include <bits/stdc++.h>using namespace std;int T;int a, b;int main(){ cin >> T; while (T--) { cin >> a >> b; cout << b - a << endl; } return 0;} B. osu!mania 从下到上，从左到右遍历字符串，输出每个\(\#\)在当前行的位置 code: 123456789101112131415161718192021222324252627#include <bits/stdc++.h>using namespace std;const int N = 510;int T;int n;string s[N];int main(){ ci ...

A. 会赢吗？ 签到，比大小 code: 123456789101112131415#include <bits/stdc++.h>using namespace std;double a, b;int main(){ cin >> a >> b; if (a < b) cout << "YES" << endl; else cout << "NO" << endl; return 0;} B. 能做到的吧 判断是否有\(i,j\)满足\(1<=i<j<=n\)，并且\(s[i]<s[j]\)，如果有就输出\(YES\)，否则输出\(NO\) code: 1234567891011121314151617181920212223242526272829303132#include <bits/stdc++.h>using namespace std;int T;s ...

A. Dora's Set 要使三个数互质，那么一定是两个奇数\(+\)一个偶数，因此遍历\(l\)到\(r\)，每次删除相邻的奇偶奇序列即可 code: 12345678910111213141516171819202122232425262728293031323334353637383940414243444546474849505152535455565758596061626364656667686970717273747576#include <bits/stdc++.h>#define int long longusing namespace std;int T;int n, m;signed main(){ cin >> T; while (T--) { cin >> n >> m; int res = 0; while (n <= m) { if (n % 2 == 0) n ...

A. Sakurako's Exam 签到，优先考虑\(2\)之间互相加减，所以只会剩下最多一个\(2\)，用\(2\)个\(1\)来消除，最后判断剩下\(1\)的个数即可 code: 1234567891011121314151617181920212223242526272829#include <bits/stdc++.h>#define int long long#define x first#define y secondusing namespace std;typedef pair<int, int> PII;const int N = 200010;int T;int a, b;signed main(){ cin >> T; while (T--) { cin >> a >> b; b %= 2; a -= b * 2; if (a < 0 || a % 2) cout << "NO&qu ...

A. Turtle and Good Strings 根据题意可知，一个好的字符串一定可以拆成两个满足条件的字符串，所以只需要令\(k=2\)进行操作即可 code: 123456789101112131415161718192021222324252627282930313233#include <bits/stdc++.h>#define int long long#define x first#define y second#define endl '\n'using namespace std;typedef pair<int, int> PII;const int N = 200010;int T;int n;string s;signed main(){ ios::sync_with_stdio(false); cin.tie(0), cout.tie(0); cin >> T; while (T--) { cin >> n >> ...

A. Primary Task 暴力判断即可 code: 1234567891011121314151617181920212223242526272829303132333435363738394041#include <bits/stdc++.h>#define x first#define y second#define int long longusing namespace std;typedef pair<int, int> PII;const int N = 200010;int T;int n;int a[N];signed main(){ cin >> T; while (T--) { cin >> n; string s; while (n > 10) { s += n % 10 + '0'; n /= 10;
...

A. Closest Point 观察一下可以发现，只有当集合中的点数等于\(2\),并且这两个点的差值大于\(1\)时，才有满足条件的点可以插入 code: 123456789101112131415161718192021222324252627282930#include <bits/stdc++.h>#define int long long#define x first#define y secondusing namespace std;typedef pair<int, int> PII;const int N = 200010;int T;int n;int a[N];signed main(){ cin >> T; while (T--) { cin >> n; for (int i = 0; i < n; i++) cin >> a[i]; if (n == 2 && abs(a[0] - a[1]) > ...

A. 日历游戏 \(sg\) 函数板子题，就是需要多处理一个日期 code: 1234567891011121314151617181920212223242526272829303132333435363738394041424344454647484950515253545556575859606162636465666768697071727374757677787980818283#include <bits/stdc++.h>using namespace std;int T;int y, m, d;int bg, ed;map<int, int> f;int mon[13] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};bool check(int y){ return (y % 4 == 0 && y % 100) || y % 400 == 0;}int get_day(int y, int m, int d){ ...

A. 该出奇兵了 $tarjan \(割点板子题，处理完割点以后\)dfs$ 求出每个点被删掉以后对答案的最小贡献 code: 123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960616263646566676869707172737475767778798081828384858687888990919293949596979899100101102103104105106107108109110111112113114115116117118119120121122123124125126127128129130#include <bits/stdc++.h>#define int __int128using namespace std;const int N = 100010;int n, m;int a[N];vector<int> h[N];int dfn[N], low[N], to ...