push_back({i, sit, c}); else if(op == 2) low ++, last = 2, low = i; else if(op == 3) up ++, last = 3, up = i; } if(last == 2) { for(int i = 1; i <= n; i ++ ) { if(s[i] >= 'A' && s[i] <= 'Z') s[i] = s[i] - 'A' + 'a'; } }...
Weak tests in D.https://atcoder.jp/contests/abc314/submissions/44532834 should fail on 7 Atcoder 3 2 0 a 1 3 a 1 3 B → Reply SummerSky 19 months ago, # | +3 Can anyone talk more about problem E? I can hardly understand the editorial. Thank you so much. → Reply judg...
} D - LOWER (abc314 D) 题目大意 给定一个字符串ss,进行三种操作: 令sx=csx=c 字符串ss全部变成小写 字符串ss全部变成大写 经过qq次操作后,输出ss。 解题思路 由于n,qn,q有2e52e5,暴力处理第二、三操作的复杂度是O(n)O(n),会超时。 操作二和操作三的影响会相互覆盖,因此我们可以只记录最后一次进行...
给定一个长度为你的由小写字母组成的字符串; 每个字符都有一个颜色; 我们让同颜色的字符首尾相连地向右移动一位; eg: abcdef的颜色为111233, 操作完之后变成了cabdfe;解题思路#开一个数组c, c[i]表示下一个颜色为i的字符要替换为c[i]; 我们把每个颜色中最靠前的字符的下标存一下, 最后更新他们即可; ...
h> using namespace std; const int N = 1e6 + 10; int res[N]; int main(){ int a, b, c, d; cin >> a >> b >> c >> d; for(int i = a; i <= b; i ++) res[i]++; for(int i = c; i <= d; i ++) res[i]++; int cnt = 0; for(int i = 0; i <= 200...
(int x,int y){if(!y)returnx;returngcd(y,x%y);}voidfj(int x){//分解质因数for(int i=2;i*i<=x;){if(!(x%i)){++t[i];//用桶记下来while(!(x%i))x/=i;}else++i;}if(x>1)++t[x];//不要忘了这一句}intmain(){scanf("%d%d",&n,&a);d=a;//d初始值为afor(int i=2...
ABC366で初めて青perfを出せました!! D問題が詰まりそうだったのでE問題を先に解いて戦略勝ちしました。 ここで遠いと思っていた水色が一気に近づきました。 そしてついにABC367で5完して入水しました!!!よっしゃーーー https://x
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AtCoder Beginner Contest 192 A~D题解 技术标签:C++算法竞赛c++算法AtCoder 文章目录 [A - Star](https://atcoder.jp/contests/abc192/tasks/abc192_a) 题目大意 输入格式 输出格式 样例 分析 代码 [B - uNrEaDaBlE sTrInG](https://atcoder.jp/contests/abc192/tasks/abc192_b)...
1449 ABC081B - Shift only C++ 17 (gcc 12.2) AtCoder *200 Aug/19/2024 22:04 1448 A - Entrance Examination C++ 17 (gcc 12.2) AtCoder *100 Aug/19/2024 22:03 1447 D - Pedometer C++ 17 (gcc 12.2) AtCoder *400 Aug/17/2024 19:04 1446 C - Enumerate Sequences C++ 17 (gcc 12.2...