这场比赛全程心态爆炸…… 开场脑子秀逗签到题WA了一发.之后0贡献. 前期状态全无 H题想复杂了,写了好久样例过不去. 然后这题还是队友过的…… 后期心态炸裂,A题后缀数组理解不深,无法特判k = 1时的情况. 然后也没有心思读题了,心静不下来. 比赛题目链接 Problem B $ans = k(n - k + 1)$ #include <bits/stdc++.h> using namespace std; typedef long long LL; LL n, k; int main()…

transaction transaction transaction Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 132768/132768 K (Java/Others)Total Submission(s): 1496    Accepted Submission(s): 723 Problem Description Kelukin is a businessman. Every day, he travels arou…

最开始一直想着最短路,不过看完题解后,才知道可以做成最长路.唉,还是太菜了. 先上图: 只要自己添加两个点,然后如此图般求最长路即可,emmm,用SPFA可以,迪杰斯特拉也可以,或者别的都ok,只要通过一次即可. 上代码: #include <cstdio> #include <cstring> #include <vector> #include <queue> #include <algorithm> using namespace std;…

题目大意: 给出两个长度为n的序列A,B,从1开始依次加Ai,减Bi,分数为第一次为当前和为负数的位置以前的Ai之和(左闭右开区间).同时有一种操作可以把当前的A1,B1移动到序列最后,注意序列A的各个元素之和等于B的各个元素之和.问取得最大分数时,至少应该操作多少次.如果分数相同,输出移动较少的次数: 基本思路: 出了以下前缀和,然后用一个sum(初始值为0)来当做判断条件,然后比这个小,就更新并记录一下下标,(每次更新说明和上一次更新这一段里的书是负数),然后最后输出下标,就是移动个数(前提…

HDU 6197 array array array 题意 输入n和k,表示输入n个整数和可以擦除的次数k,如果至多擦除k次能是的数组中的序列是不上升或者是不下降序列,就是魔力数组,否则不是. 解题思路 分别求最长不下降和不上升子序列的长度,看不符合要求的数字和k的大小. 这里使用优化后的求解最长不上升和不下降子序列的算法. #include <cstdio> #include <algorithm> using namespace std; ; ; int A[maxn], B[…

题目链接 emmmm...思路是群里群巨聊天讲这题是用尺取法.....emmm然后就没难度了,不过时间上3000多,有点.....盗了个低配本的读入挂发现就降到2800左右, 翻了下,发现神犇Claris280MS秒过.......%%% #include <stdio.h> #include <stdlib.h> #include <cmath> #include <string.h> #include <iostream> #include…

思路:找规律发现这个数是斐波那契第2*k+3项-1,数据较大矩阵快速幂搞定.   快速幂入门第一题QAQ #include <stdio.h> #include <stdlib.h> #include <cmath> #include <string.h> #include <iostream> #include <algorithm> #include <queue> #include <vector> #i…

2017-09-15 21:05:41 writer:pprp 给出一个序列问能否去掉k的数之后使得整个序列不是递增也不是递减的 先求出LIS,然后倒序求出最长递减子序列长度,然后判断去k的数后长度是否都大于所求长度 代码如下: #include <bits/stdc++.h> using namespace std; ],tmp1[],arr2[], tmp2[]; int len1,len2; int main() { int cas; // cin >> cas; scanf…

题意:看后面也应该知道是什么意思了 解法: 我们设置l,r,符合条件就是l=起始点,r=当前点,不符合l=i+1 学习了一下FASTIO #include <iostream> #include <algorithm> #include <set> #include <string> #include <vector> #include <queue> #include <map> #include <stack&g…

Problem Description Kelukin is a businessman. Every day, he travels around cities to do some business. On August 17th, in memory of a great man, citizens will read a book named "the Man Who Changed China". Of course, Kelukin wouldn't miss this c…

题意:求n个斐波那契数列组合都无法得到的最小数字 解法: 1 我们先暴力的求出前面几个数字 2 然后再暴力的求递推 3 接着矩阵快速幂(没写错吧?) /*#include<bits/stdc++.h> using namespace std; long long x[50]; map<long long,int>Mp; void dfs(int pos,long long sum,int cnt,int n){ if(cnt==n){ Mp[sum]=1; return; } if…

Problem Description Connecting the display screen and signal sources which produce different color signals by cables, then the display screen can show the color of the signal source.Notice that every signal source can only send signals to one display…

cable cable cable Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2084    Accepted Submission(s): 1348 Problem Description Connecting the display screen and signal sources which produce differen…

题目链接  Guangxi 感觉这场比赛完全是读题场啊…… 比赛过程中丢失了一波进度,最后想开题的时候已经来不及了…… Problem A 按题意模拟……按照那个矩阵算就可以了 #include <bits/stdc++.h> using namespace std; #define rep(i, a, b) for (int i(a); i <= (b); ++i) #define dec(i, a, b) for (int i(a); i >= (b); --i) double…

题目链接  Beijing…

题目链接  Xian…

题目链接  Qingdao Problem C AC自动机还不会,暂时暴力水过. #include <bits/stdc++.h> using namespace std; #define rep(i, a, b) for (int i(a); i <= (b); ++i) #define dec(i, a, b) for (int i(a); i >= (b); --i) const int N = 1e5 + 10; string s[N]; int T; int n; int…

比赛题目链接  Urumuqi…

最开始一直想不通,为什么推出这个公式,后来想了半天,终于想明白了. 题目大意是,有M个格子,有K个物品.我们希望在格子与物品之间连数量尽可能少的边,使得——不论是选出M个格子中的哪K个,都可以与K个物品恰好一一匹配. 然后你可以试着画图,每次必须有k个格子是单独的(与各物体只有一条线相连)所以还剩下m-k个格子,可以用来补位,也就是跟每个物品都相连,所以就有(m-k)*k 上代码(巨巨巨巨巨简单): #include <cstdio> #include <cstring> #inc…

分析: 当n=1时ans=4=f(5)-1; n=2,ans=12=f(7)-1; n=3,ans=33=f(9)-1; 于是大胆猜想ans=f(2*k+3)-1. 之后用矩阵快速幂求解f(n)即可,O(logn). AC code: #include<bits/stdc++.h> using namespace std; typedef long long ll; typedef vector<ll> vec; typedef vector<vec> mat; ; m…

Problem Description We define a sequence F: ⋅ F0=0,F1=1;⋅ Fn=Fn−1+Fn−2 (n≥2). Give you an integer k, if a positive number n can be expressed byn=Fa1+Fa2+...+Fak where 0≤a1≤a2≤⋯≤ak, this positive number is mjf−good. Otherwise, this positive number is …

hdu 6197 题意:给定一个数组,问删掉k个字符后数组是否能不减或者不增,满足要求则是magic array,否则不是. 题解:队友想的思路,感觉非常棒!既然删掉k个后不增或者不减,那么就先求数组的最长不下降子序列的长度l1和最长不上升子序列的长度l2,若l1>=n-k||l2>=n-k,则满足要求. #include<iostream> #include<cstdio> #include<cstring> #include<algorithm&g…

网络赛:2017 ACM/ICPC Asia Regional Shenyang Online 题目来源:cable cable cable Problem Description: Connecting the display screen and signal sources which produce different color signals by cables, then the display screen can show the color of the signal sou…

2017 ACM ICPC Asia Regional - Daejeon Problem A Broadcast Stations 题目描述:给出一棵树,每一个点有一个辐射距离\(p_i\)(待确定),但\(p_i==0\)的点不能辐射自己,只能由别的点辐射覆盖.求\(p_i\)的和的最小值. Problem B Connect3 题目描述:有一个\(4 \times 4\)的网格,两个人玩游戏,第一个人用黑棋,第二人用白棋.每次选择一列,将棋子扔下去,直到最下一个空的格子.现在给出第一个人下…

hannnnah_j’s Biological Test Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 802    Accepted Submission(s): 269 Problem Description hannnnah_j is a teacher in WL High school who teaches biolog…

QSC and Master Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 859    Accepted Submission(s): 325 Problem Description Every school has some legends, Northeastern University is the same. Enter…

odd-even number Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 388    Accepted Submission(s): 212 Problem Description For a number,if the length of continuous odd digits is even and the length…

Apple Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)Total Submission(s): 982    Accepted Submission(s): 323 Problem Description Apple is Taotao's favourite fruit. In his backyard, there are three apple trees with…

Brute Force Sorting Time Limit: 1 Sec  Memory Limit: 128 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=6215 Description Beerus needs to sort an array of N integers. Algorithms are not Beerus's strength. Destruction is what he excels. He can destr…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5458 Problem Description Given an undirected connected graph G with n nodes and m edges, with possibly repeated edges and/or loops. The stability of connectedness between node u and node v is defined by…