简单判断. 分别判断每个单词是几面体,加起来就是答案. #include <cstdio> #include <cmath> #include <cstring> #include <map> #include <algorithm> using namespace std; int n; char s[1000]; int sum; int main() { scanf("%d",&n); for(int i=1;i…

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]:http://codeforces.com/contest/785 [题意] 给你各种形状的物体; 然后让你计算总的面数; [题解] 用map来记录各种物体本该有的面数; 读入各种物体; 然后累加各种物体的面数; 然后输出就好; [完整代码] #include <bits/stdc++.h> using namespace std; #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define LL l…

E. Anton and Tree time limit per test: 3 seconds memory limit per test :256 megabytes input:standard input output: standard output Anton is growing a tree in his garden. In case you forgot, the tree is a connected acyclic undirected graph. There are …

LINK time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output The teacher gave Anton a large geometry homework, but he didn't do it (as usual) as he participated in a regular round on Codeforces. In…

[题目链接] http://codeforces.com/problemset/problem/734/F [题目大意] 给出数列b和数列c,求数列a,如果不存在则输出-1 [题解] 我们发现: bi+ci=2n*ai-(所有ai为1且aj为0的数位)+(ai为0且aj为1的数位)= n*ai+Σak 记为(1)式 同时又有(ai为1且aj为0的数位)+(aj为1且ai为0的数位)=ai xor aj 那么Σ(bi+ci)=Σ(2n*ai)-Σ(ai xor aj)+Σ(ai xor aj)=Σ…

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 different sequences that consist of round brackets (characters "(" and &quo…

题目链接:https://codeforces.com/problemset/problem/785/D 题解: 首先很好想的,如果我们预处理出每个 "(" 的左边还有 $x$ 个 "(",以及右边有 $y$ 个 ")",那么就有式子如下: ① 若 $x+1 \le y$:$C_{x}^{0} C_{y}^{1} + C_{x}^{1} C_{y}^{2} + \cdots + C_{x}^{x} C_{y}^{x+1} = \sum_{i=0}…

题目链接:https://codeforces.com/contest/584/problem/E 题意: 给两个 $1 \sim n$ 的排列 $p,s$,交换 $p_i,p_j$ 两个元素需要花费 $|i-j|$,要求你用最少的钱交换 $p$ 中的元素使得其变成 $s$. 题解: 思路很简单,给个例子如下: $p$:$5,1,2,3,4,6$ $s$:$3,4,1,6,2,5$ 我们不难发现,像: $p$:$5,1,\underline{2},\underline{3},4,6$ $s$:$…

题目链接:http://codeforces.com/problemset/problem/785/E 其实可以CDQ分治... 我们只要用一个数据结构支持单点修改,区间查询比一个数大(小)的数字有多少个就可以了. 考虑分块,每段区间之内有排序或者二分查询比一个数大的树的个数的操作. 复杂度${O(qn \sqrt n log_{2}^{\sqrt n})}$ #include<iostream> #include<cstdio> #include<algorithm>…