http://codeforces.com/blog/entry/50996 官方题解讲得很明白,在这里我复述一下. 枚举每个左括号,考虑计算一定包含其的简单括号序列的个数,只考虑其及其左侧的左括号,以及其右侧的右括号.最后答案就是其之和. 可以将其提取出来这样((((((())),红色为当前左括号.设有x个左,y个右 要注意,这个答案为C(x+y-1,x),来证明. 我们只需证明,这个答案与长度为x+y-1的,包含x个1的零一序列的种类数相等即可. 随便写一个这样的零一序列,长度为x+y,但当…

D. Anton and School - 2 题目连接: http://codeforces.com/contest/785/problem/D Description As you probably know, Anton goes to school. One of the school subjects that Anton studies is Bracketology. On the Bracketology lessons students usually learn differ…

C. Anton and Fairy Tale 题目连接: http://codeforces.com/contest/785/problem/C Description Anton likes to listen to fairy tales, especially when Danik, Anton's best friend, tells them. Right now Danik tells Anton a fairy tale: "Once upon a time, there liv…

B. Anton and Classes 题目连接: http://codeforces.com/contest/785/problem/B Description Anton likes to play chess. Also he likes to do programming. No wonder that he decided to attend chess classes and programming classes. Anton has n variants when he wil…

A - Anton and Polyhedrons 题目连接: http://codeforces.com/contest/785/problem/A Description Anton's favourite geometric figures are regular polyhedrons. Note that there are five kinds of regular polyhedrons: Tetrahedron. Tetrahedron has 4 triangular face…

题目链接 转自 给你一个字符串问你能构造多少RSBS. #include<bits/stdc++.h> #define LL long long #define fi first #define se second #define mp make_pair #define pb push_back using namespace std; LL gcd(LL a,LL b){return b?gcd(b,a%b):a;} LL lcm(LL a,LL b){return a/gcd(a,b)*…

当m>=n时,显然答案是n: 若m<n,在第m天之后,每天粮仓减少的量会形成等差数列,只需要二分到底在第几天,粮仓第一次下降到0即可. 若直接解不等式,可能会有误差,需要在答案旁边扫一下. 注意二分上界的确定,不能太小也不能太大. #include<cstdio> #include<iostream> using namespace std; typedef long long ll; ll n,m; int main(){ cin>>n>>m;…

E. Anton and Permutation time limit per test 4 seconds memory limit per test 512 megabytes input standard input output standard output Anton likes permutations, especially he likes to permute their elements. Note that a permutation of n elements is a…

Codeforces Round #404 (Div. 2) 题意:对于 n and m (1 ≤ n, m ≤ 10^18)  找到 1) [n<= m] cout<<n; 2) [n>m]最小的 k => (k -m) * (k-m+1) >= (n-m)*2 成立 思路:二分搜索 #include <bits/stdc++.h> #include <map> using namespace std; #define LL long long…

题目传送门 /* 题意:从前面找一个数字和末尾数字调换使得变成偶数且为最大 贪心:考虑两种情况:1. 有偶数且比末尾数字大(flag标记):2. 有偶数但都比末尾数字小(x位置标记) 仿照别人写的,再看自己的代码发现有清晰的思维是多重要 */ #include <cstdio> #include <iostream> #include <algorithm> #include <cmath> #include <cstring> #include…