B. Far Relative's Problem 题目连接: http://www.codeforces.com/contest/629/problem/B Description Famil Door wants to celebrate his birthday with his friends from Far Far Away. He has n friends and each of them can come to the party in a specific range of…

题意:n个人,在规定时间范围内,找到最多有多少对男女能一起出面. 思路:ans=max(2*min(一天中有多少个人能出面)) #include<iostream> #include<string> #include<algorithm> #include<cstdlib> #include<cstdio> #include<set> #include<map> #include<vector> #includ…

A. Far Relative's Birthday Cake 题目连接: http://www.codeforces.com/contest/629/problem/A Description Door's family is going celebrate Famil Doors's birthday party. They love Famil Door so they are planning to make his birthday cake weird! The cake is a…

A. Far Relative's Birthday Cake time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Door's family is going celebrate Famil Doors's birthday party. They love Famil Door so they are planning to m…

A. Far Relative’s Birthday Cake time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Door's family is going celebrate Famil Doors's birthday party. They love Famil Door so they are planning to m…

水题 #include<iostream> #include<string> #include<algorithm> #include<cstdlib> #include<cstdio> #include<set> #include<map> #include<vector> #include<cstring> #include<stack> #include<cmath>…

B. Far Relative’s Problem time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Famil Door wants to celebrate his birthday with his friends from Far Far Away. He has n friends and each of them c…

居然补完了 组合 A - Far Relative’s Birthday Cake import java.util.*; import java.io.*; public class Main { public static void main(String[] args) { Scanner cin = new Scanner (new BufferedInputStream (System.in)); int n = cin.nextInt (); int[] col = new int[…

A. Far Relative's Birthday Cake 题意: 求在同一行.同一列的巧克力对数. 分析: 水题~样例搞明白再下笔! 代码: #include<iostream> using namespace std; const int maxn = 105; char a[maxn][maxn]; int main (void) { int N;cin>>N; int cnt = 0, res = 0; for(int i = 0; i < N; i++){ cn…

E. Famil Door and Roads 题目连接: http://www.codeforces.com/contest/629/problem/E Description Famil Door's City map looks like a tree (undirected connected acyclic graph) so other people call it Treeland. There are n intersections in the city connected b…

D. Babaei and Birthday Cake 题目连接: http://www.codeforces.com/contest/629/problem/D Description As you know, every birthday party has a cake! This time, Babaei is going to prepare the very special birthday party's cake. Simple cake is a cylinder of som…

C. Famil Door and Brackets 题目连接: http://www.codeforces.com/contest/629/problem/C Description As Famil Door's birthday is coming, some of his friends (like Gabi) decided to buy a present for him. His friends are going to buy a string consisted of roun…

题目链接: http://www.codeforces.com/contest/629/problem/E 题解: 树形dp. siz[x]为x这颗子树的节点个数(包括x自己) dep[x]表示x这个节点的深度,从1开始(其实从什么开始都可以,我们这里用到的只是相对距离) 对于查询u,v,总共有三种情况: 1.u为公共祖先 设x为(u,v)链上u的儿子,则我们知道新边只能从非x子树的点(n-siz[x]连到以v为根的子树上的点(siz[v]) 则新边的总条数为(n-siz[x])*siz[v]…

题目链接: http://codeforces.com/contest/629/problem/C 题意: 长度为n的括号,已经知道的部分的长度为m,现在其前面和后面补充‘(',或')',使得其长度为n,且每个左括号都能找到右括号与它匹配. 题解: dp[i][j]表示长度为i,平衡度为j的合法括号序列的总数,这里平衡度定义是‘('比')'多多少个,或')'比’('多多少个. #include<iostream> #include<cstring> #include<cstd…

A. Far Relative’s Birthday Cake time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Door's family is going celebrate Famil Doors's birthday party. They love Famil Door so they are planning to m…

Famil Door's City map looks like a tree (undirected connected acyclic graph) so other people call it Treeland. There are n intersections in the city connected by n - 1 bidirectional roads. There are m friends of Famil Door living in the city. The i-t…

题意:做蛋糕,给出N个半径,和高的圆柱,要求后面的体积比前面大的可以堆在前一个的上面,求最大的体积和. 思路:首先离散化蛋糕体积,以蛋糕数量建树建树,每个节点维护最大值,也就是假如节点i放在最上层情况下的体积最大值dp[i].每次查询比蛋糕i小且最大体积的蛋糕,然后更新线段树.注意此题查询的技巧!!查询区间不变l,r,才能保证每次查到的是小且最大体积. #include<iostream> #include<string> #include<algorithm> #in…

题目连接:http://codeforces.com/contest/798/problem/C C. Mike and gcd problem time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Mike has a sequence A = [a1, a2, ..., an] of length n. He considers…

E. Mike and Geometry Problem 题目连接: http://www.codeforces.com/contest/689/problem/E Description Mike wants to prepare for IMO but he doesn't know geometry, so his teacher gave him an interesting geometry problem. Let's define f([l, r]) = r - l + 1 to…

任意门:http://codeforces.com/contest/689/problem/E E. Mike and Geometry Problem time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard output Mike wants to prepare for IMO but he doesn't know geometry, so…

题目链接:http://codeforces.com/contest/742/problem/B B. Arpa's obvious problem and Mehrdad's terrible solution time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output There are some beautiful girls in…

                                                                       C. A Problem about Polyline                                                                                time limit per test 1 second                                            …

B. Arpa’s obvious problem and Mehrdad’s terrible solution time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output There are some beautiful girls in Arpa’s land as mentioned before. Once Arpa came u…

E. Little Girl and Problem on Trees time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output A little girl loves problems on trees very much. Here's one of them. A tree is an undirected connected g…

E. Mike and Geometry Problem time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard output Mike wants to prepare for IMO but he doesn't know geometry, so his teacher gave him an interesting geometry pro…

如果一开始就满足题意,不用变换. 否则,如果对一对ai,ai+1用此变换,设新的gcd为d,则有(ai - ai+1)mod d = 0,(ai + ai+1)mod d = 0 变化一下就是2 ai mod d = 0 2 ai+1 mod d = 0 也就是说,用两次变换之后,gcd至少扩大2倍,于是,最优方案就是我们将所有的奇数都变成偶数. 只需要找出所有奇数段,答案就是sigma([奇数段的长度/2]+(奇数段的长度 mod 2 ==1 ?)). #include<cstdio> #i…

题目链接:传送门 题目大意:给你n个区间,求任意k个区间交所包含点的数目之和. 题目思路:将n个区间都离散化掉,然后对于一个覆盖的区间,如果覆盖数cnt>=k,则数目应该加上 区间长度*(cnt与k的组合数) ans=ans+(len*C(cnt,k))%mod; #include <iostream> #include <cstdio> #include <cstdlib> #include <cmath> #include <algorith…

题目中给出的函数具有周期性,总可以移动到第一个周期内,当然,a<b则无解. 假设移动后在上升的那段,则有a-2*x*n=b,注意限制条件x≥b,n是整数,则n≤(a-b)/(2*b).满足条件的x≥(a-b)/(2*n) 假设在下降的那段,2*x-(a-2*x*n)=b,n+1≤(a+b)/(2*b),x≥(a+b)/(2*(n+1)) 两者取最小值 #include<bits/stdc++.h> using namespace std; int main() { int a,b; sc…

[链接]h在这里写链接 [题意]     给你一个金额N,和硬币的类型总数M;     (完全背包),然后问你组成N的方案数.     使得,用这些硬币组成价值为N的金额的方案数为A;     现在A已知,让你求出一组合法的N,M以及每个硬币的面值. [题解] 只要面值为1和面值为2. 做个完全背包.就能发现这两个能够覆盖到所有的方案数. [错的次数] 0 [反思] 在这了写反思 [代码] #include <bits/stdc++.h> using namespace std; long l…

yyb大佬的博客 这线段期望好神啊... 还有O(nlogn)FFTO(nlogn)FFTO(nlogn)FFT的做法 Freopen大佬的博客 本蒟蒻只会O(n2)O(n^2)O(n2) CODE #include <bits/stdc++.h> using namespace std; const int mod = 998244353; typedef long long LL; const int MAXN = 4005; inline void add(int &x, int…