AC，∑ndless

以下内容基于 CF896C Willem, Chtholly and Seniorious 来介绍。珂朵莉树实质上是一种可以维护区间上的分裂与合并的数据结构，但要求数据是随机的，或者有大量的随机合并操作，这样才能保证维护的区间个数是一个很小的值。一开始，我们用不同的节点表示 [1,1],[2,2],...,[n,n][1,1],[2,2],...,[n,n][1,1],[2,2],...,[n...

https://codeforces.com/problemset/problem/1251/D提供一个比官方题解还简洁的二分思路。首先，中位数越大，发出的工资越多，这满足二分条件。于是二分中位数，判断所发工资是否不超过 sss，整个过程如下：二分前预处理，将数据按照 lll 从大到小排序，另外还需计算基础工资之和 baseCostbaseCostbaseCost，即所有 lll 之和...

A. Secretshttp://codeforces.com/contest/333/problem/Atime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputGe

http://codeforces.com/contest/140/problem/A过大圆圆心作小圆切线即可发现规律，详见代码。注意判相等一定要用fabs！！！完整代码：/*30ms,0KB*/#includeusing namespace std;int main(){ int n, R, r; double a; scanf("%d%d%d"

A. Cut Ribbonhttp://codeforces.com/problemset/problem/189/Atime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard output

B. Spreadsheetshttp://codeforces.com/problemset/problem/1/Btime limit per test10 secondsmemory limit per test64 megabytesinputstandard inputoutputstandard output

http://codeforces.com/contest/371/problem/C直接二分能做的汉堡数，注意上界要开大一点。完整代码：/*15ms,0KB*/#includeusing namespace std;const long long mx = 1e13;///多开一位，因为结果可以略大于1e12const char b[] = "BSC";

http://codeforces.com/contest/371/problem/B神题必有神解——你能想到这么做吗？/*15ms,0KB*/#includeusing namespace std;int op_cnt[6];int main(){ int a, b, ans = 0; cin >> a >> b; for (int n = 2; n <=

http://codeforces.com/contest/190/problem/D【神题必有神解】从这一题可大题了解two pointers算法的威力。/*1024ms,42062KB*/#includeusing namespace std;const int mx = 400005;int a[mx];map mp;/*数据例子：10 22 3

http://codeforces.com/problemset/problem/359/D/*124ms,2300KB*/#includeint a[300005], w[300005];int main(){ int n, i, l, r, cnt = 0, maxd = 0; scanf("%d", &n); for (i = 0; i < n; i++) s

C. Find Maximumhttp://codeforces.com/problemset/problem/353/Ctime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard output

A. Polo the Penguin and Stringshttp://codeforces.com/problemset/problem/288/Atime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutput

A. Insomnia curehttp://codeforces.com/problemset/problem/148/Atime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard output

B. Binary Numberhttp://codeforces.com/problemset/problem/92/Btime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard output

A. Puzzleshttp://codeforces.com/contest/337/problem/Atime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputTh

B. Routine Problemhttp://codeforces.com/contest/337/problem/Btime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard output

A. Extra-terrestrial Intelligencehttp://codeforces.com/problemset/problem/36/Atime limit per test2 secondsmemory limit per test64 megabytesinputinput.txtoutputoutpu

B. Chilly Willyhttp://codeforces.com/problemset/problem/248/Btime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard output

A. Free Cashhttp://codeforces.com/problemset/problem/237/Atime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard output

A. System of Equationstime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputFurik loves math lessons very much,

A. Perfect Pairhttp://codeforces.com/problemset/problem/317/Atime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard output

http://codeforces.com/problemset/problem/401/D解释全部在代码的注释中：/*78ms,205464KB*/#includeusing namespace std;const int mx = 1 << 18;long long dp[mx][100];///dp[mask][j]表示余数为j时的mask对应的x的个数///

http://codeforces.com/contest/337/problem/D/*154ms,10300KB*/#includeusing namespace std;const int mx = 100005;vector g[mx];int d1[mx], d2[mx], p[mx];void dfs(int p, int u, int deep, i

http://codeforces.com/problemset/problem/116/C从树根DFS，看最大能递归几层。/*30ms,300KB*/#includeusing namespace std;const int mx = 2005;vector v[mx];bool fa[mx];int maxlen;void dfs(int i, int

http://codeforces.com/problemset/problem/303/Awhen n is odd, A[i] = B[i] = iwhen n is even, there is no solution.why? If , then  or just , where S = 0 + 1 + ... + n - 1 = n(n - 1) / 

http://codeforces.com/problemset/problem/311/AIf we ignore "break", tot will be up to .Consider whether we can make such inequality d ≤ p[j].x - p[i].x is always false. The obvious way

A. The Wallhttp://codeforces.com/contest/340/problem/Atime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputI

http://codeforces.com/contest/404/problem/C思路：我们构造一颗最短路径树就行了。若能够构造，边数必然为n-1（样例1的边数可以是两条）。如何构造？从距离为1的点开始，逐渐往下加边，生成一颗k叉树。若在中间生成了大于k的叉，则输出-1。完整代码：/*265ms,4436KB*/#includeusing namesp

http://codeforces.com/contest/387/problem/C/*31ms,100KB*/#includechar str[100005];int main(){ gets(str); int i, j, res = 0; for (i = 0; str[i]; i = j) { for (j = i + 1; str[j] == '0

http://codeforces.com/contest/404/problem/B/*124ms,0KB*/#includeusing namespace std;double a;void f(double len){ int c = ((int)(len / a)) % 4; len = fmod(len, a); if (c == 0) printf("

先预处理下：在每个字符的两边都插入一个特殊的符号，比如abba变成#a#b#b#a#，aba变成#a#b#a#（因为Manacher算法只能处理奇数长度的字符串）。同时，为了避免数组越界，在字符串开头添加另一特殊符号，比如$#a#b#a#。以字符串3212343219为例，处理后变成S[] = "$#3#2#1#2#3#4#3#2#1#9#"。然后用一个数组Len[i]来记录以字符S[i

http://codeforces.com/contest/370/problem/C/*31ms,0KB*/#include#includeusing namespace std;int C[5005];int main(){ int N, M; cin >> N >> M; for (int i = 0; i > C[i]; sort(C, C + N

http://codeforces.com/contest/339/problem/C爆搜水过，复杂度貌似是O(m)的？/*62ms,4KBm*/#includeint m, ans[1005]; ///ans从1开始char s[15];bool ok;///O(m)复杂度？int dfs(int deep, int diff) /// diff表示重量之差{

http://codeforces.com/problemset/problem/116/D/*30ms,0KB*/#includeconst int mx = 155;char g[mx][mx];int left[mx], right[mx];int main(){ int n, m, i, j, posy, dir; int cnt = -1; scanf

http://codeforces.com/problemset/problem/245/D/*30ms,0KB*/#includeusing namespace std;const int mx = 105;int m[mx][mx],a[mx];int main(){ int n, i, j; scanf("%d", &n); for (i = 1; i

http://codeforces.com/problemset/problem/245/F#includeusing namespace std;const int mx = 5000005;const int month[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};const int day2s = 24

http://codeforces.com/contest/340/problem/B枚举一条线段，再枚举这条线段左右两边的三角形，拼起来就是一个四边形了。完整代码：/*186ms,0KB*/#includeusing namespace std;int x[300], y[300], A[2];int main(){ int n, i, j, k,

http://codeforces.com/contest/370/problem/B/*15ms,200KB*/#includeusing namespace std;vector v[105], tmp(105);int main(){ int n, i, j, k, x; scanf("%d", &n); for (i = 0; i < n; ++i) {

http://codeforces.com/contest/355/problem/C枚举i，左手操作了i次，右手操作了n-i次，然后重复次数可以直接算出来，所以答案就可以在O(n)的时间内算出来。/*30ms,4000KB*/#includeusing namespace std;int a[100005];int main(){ int n, l, r,

http://codeforces.com/problemset/problem/116/B开始还打算打匈牙利的。。结果看到了这句话：“there will be at most one wolf adjacent to each little pig”也就是说，对于每头狼，若它周围有猪，就++cnt因为不会出现两头狼吃同一只猪的情况。完整代码：/*30ms