意甲冠军:给定的长度可达1000数的顺序,图像password像锁.可以上下滑动,同时会0-9周期. 每个操作.最多三个数字连续操作.现在给出的起始序列和靶序列,获得操作的最小数量,从起始序列与靶序列. 花了一天的时间.我觉得是道非常难的DP.这个阶段非常好划分,对于前面完毕的password锁就不再考虑.问题的关键是这个旋转每次能够的情况非常多. 同一时候也能够发现当I位置上确定移好后,至多影响到后两位.-> dp[i][j][k] 表示当前i位移好后,i+1 为j ,  i+2为k 的次数.…
Sum of divisors Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4318    Accepted Submission(s): 1382 Problem Description mmm is learning division, she's so proud of herself that she can figure…
Problem Description A password locker with N digits, each digit can be rotated to 0-9 circularly.You can rotate 1-3 consecutive digits up or down in one step.For examples:567890 -> 567901 (by rotating the last 3 digits up)000000 -> 000900 (by rotati…
Problem Description In this problem, you are given several strings that contain only digits from '0' to '9', inclusive.An example is shown below.101123The set S of strings is consists of the N strings given in the input file, and all the possible sub…
Problem Description There's a queue obeying the first in first out rule. Each time you can either push a number into the queue (+i), or pop a number out from the queue (-i). After a series of operation, you get a sequence (e.g. +1 -1 +2 +4 -2 -4). We…
Problem Description Japanese Mahjong is a four-player game. The game needs four people to sit around a desk and play with a set of Mahjong tiles. A set of Mahjong tiles contains four copies of the tiles described next:One to nine Man, which we use 1m…
GCC Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 3867    Accepted Submission(s): 1272 Problem Description The GNU Compiler Collection (usually shortened to GCC) is a compiler system produc…
感慨一下,区域赛的题目果然很费脑啊!!不过确实是一道不可多得的好题目!! 题目大意:给你一棵有n个节点的树,让你移动树中一条边的位置,即将这条边连接到任意两个顶点(边的大小不变),要求使得到的新树的直径最小. 解题思路:此题先求出原始树的直径maxr1,并记录直径上的各个节点.很容易想到要移动的边一定是直径上的边,只有这样才有可能使树的直径减小!! 接着就是枚举直径上的每条边,并用这条边作为分隔将原始树分割成两棵子树(即子树一和子树二),然后分别求子树一的直径maxr2 和子树二的直径maxr3…
Description You are given an undirected graph with N vertexes and M edges. Every vertex in this graph has an integer value assigned to it at the beginning. You're also given a sequence of operations and you need to process them as requested. Here's a…
Description “Be subtle! Be subtle! And use your spies for every kind of business. ”― Sun Tzu“A spy with insufficient ability really sucks”― An anonymous general who lost the warYou, a general, following Sun Tzu’s instruction, make heavy use of spies…
Description P. T. Tigris is a student currently studying graph theory. One day, when he was studying hard, GS appeared around the corner shyly and came up with a problem: Given a graph with n nodes and m undirected weighted edges, every node having o…
Browsing History http://acm.hdu.edu.cn/showproblem.php?pid=4464 签到 #include<cstdio> #include<algorithm> using namespace std; ]; int main(){ ; while(~scanf("%d",&n)){ ; while(n--){ scanf("%s",a); ; ;a[i];i++){ sum+=a[i];…
Friend Chains http://acm.hdu.edu.cn/showproblem.php?pid=4460 图的最远两点距离,任意选个点bfs,如果有不能到的点直接-1.然后对于所有距离最远的点都bfs一次.最坏n^2 邻接表 #include<cstdio> #include<cstring> #include<iostream> #include<queue> #include<map> #define mt(a,b) mems…
Draw Something http://acm.hdu.edu.cn/showproblem.php?pid=4450 o(n)统计输入每个数的平方和. #include<cstdio> int main(){ int n,x; while(~scanf("%d",&n),n){ ; while(n--){ scanf("%d",&x); ans+=x*x; } printf("%d\n",ans); } ; }…
标题效果 有着n巫妖.m精灵.k木.他们都有自己的位置坐标表示.冷却时间,树有覆盖范围. 假设某个巫妖攻击精灵的路线(他俩之间的连线)经过树的覆盖范围,表示精灵被树挡住巫妖攻击不到.求巫妖杀死所有精灵的时间.若无法所有杀死输出-1: 解题思路: 推断巫妖能否打到精灵用线段与点的最短距离来推断,若最短距离小于树的覆盖范围,就攻击不到. 最小时间能够跑费用流来解决,也能够二分图的最优匹配. 以下是代码: #include <set> #include <map> #include &l…
推公式 #include <cstdio> #include <cmath> #include <iomanip> #include <iostream> #include <algorithm> using namespace std; typedef long long ll; const int N = 50000+3; ll a[N], b[N]; int main() { int T, n, m, len; ll x, sum, ans…
Contest Problem Description In the ACM International Collegiate Programming Contest, each team consist of three students. And the teams are given 5 hours to solve between 8 and 12 programming problems. On Mars, there is programming contest, too. Each…
4576 njczy2010 C Accepted 860 KB 140 ms G++ 2063 B 2014-10-16 09:51:19 哎,为啥1000*100*100的复杂度的dp就不敢敲了呢,,,真是2 涉及到可能有后效性的时候,一维就不行了,要扩维.本题,一个状态的变化会影响后两个,所以要用三维. lockerTime Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Tota…
Conquer a New Region Time Limit: 5 Seconds      Memory Limit: 32768 KB The wheel of the history rolling forward, our king conquered a new region in a distant continent. There are N towns (numbered from 1 to N) in this region connected by several road…
WHUgirls Time Limit: 3000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 2068    Accepted Submission(s): 785 Problem Description There are many pretty girls in Wuhan University, and as we know, every girl lo…
摘要 本文主要给出了2018 ACM-ICPC Asia Beijing Regional Contest的部分题解,意即熟悉区域赛题型,保持比赛感觉. Jin Yong’s Wukong Ranking List 题意 输入关系组数n和n组关系,每组关系是s1 > s2,问第一出现矛盾的组,或者没有矛盾就输出0. 解题思路 第一感觉是拓扑排序,未完,又写了一个深搜的传递闭包,1 A,和2018年河南省赛的题很像. 代码 #include <cstdio> #include <ma…
摘要 本文主要给出了2014-2015 ACM-ICPC, Asia Xian Regional Contest的部分题解,说明了每题的题意.解题思路和代码实现,意即熟悉区域赛比赛题型. Built with Qinghuai and Ari Factor 题意 判断是否是Q数列,只要数列中每个数均能够被3整除就是Q数列. 解题思路 需要特判一下0的情况. 代码 #include <cstdio> int main() { int T; int n; ; scanf("%d"…
摘要: 本文是The 2018 ACM-ICPC Asia Qingdao Regional Contest(青岛现场赛)的部分解题报告,给出了出题率较高的几道题的题解,希望熟悉区域赛的题型,进而对其他区域赛的准备有借鉴意义. Function and Function 题意 给出x和k,计算gk(x). 解题思路 通过观察发现,g函数经过一定次数的递推一定会在0和1之间变换,所以循环内加判断提前结束递推即可. 易错分析 注意计算f(0)返回的是1的问题,下面的写法避免了这种错误. 代码实现 #…
The 2018 ACM-ICPC Asia Qingdao Regional Contest 青岛总体来说只会3题 C #include<bits/stdc++.h> using namespace std; #define maxn 3000005 char a[maxn],b[maxn]; int c[maxn],ll[maxn],rr[maxn]; int main(){ int t; cin>>t; while(t--){ int n; scanf("%d&qu…
2017-2018 ACM-ICPC, Asia Tsukuba Regional Contest A Secret of Chocolate Poles 思路:暴力枚举黑巧克力的个数和厚黑巧克力的个数 代码: #pragma GCC optimize(2) #pragma GCC optimize(3) #pragma GCC optimize(4) #include<bits/stdc++.h> using namespace std; #define fi first #define s…
The 2014 ACM-ICPC Asia Mudanjiang Regional Contest A.Average Score B.Building Fire Stations C.Card Game D.Domination E.Excavator Contest F.Fiber-optic Network G.Garden and Sprinklers H.Hierarchical Notation I.Information Entropy J.Jacobi Pattern K.Kn…
The 2014 ACM-ICPC Asia Mudanjiang Regional Contest 题目链接 没去现场.做的网络同步赛.感觉还能够,搞了6题 A:这是签到题,对于A堆除掉.假设没剩余在减一.B堆直接除掉 + 1就能够了 B:二分贪心,二分长度.然后会发现本质上是在树上最长链上找两点,那么有二分出来的长度了,就从两端分别往里移动那么长,那两个位置就是放置位置.然后在推断一下就能够了 D:概率DP.首先知道放一个棋子.能够等价移动到右上角区域,那么就能够dp[x][y][k],表示…
2014-2015 ACM-ICPC, Asia Tokyo Regional Contest A B C D E F G H I J K O O O O   O O         A - Bit String Reordering 签到 #include <bits/stdc++.h> using namespace std; ; ]; int temp[N]; int put(int opt) { ] = {}; ; ; i <= m; i++) { int num = p[i];…
2019-2020 ICPC, Asia Jakarta Regional Contest (Online Mirror, ICPC Rules, Teams Preferred) easy: ACEGHK medium-easy: BJL medium: D ?????: I A. B. C. 对 \(R[],C[]\) 分别按奇偶性分段. 网! D. 考虑 check 一个串,枚举右走对应的前缀 pre,下走对应的后缀 suf. 把每行反串拼接中间连特殊字符,建 SA,能 match 上 p…
ACM-ICPC Asia Beijing Regional Contest 2018 Reproduction hihocoder1870~1879 A 签到,dfs 或者 floyd 都行. #include<bits/stdc++.h> using namespace std; typedef long long LL; typedef long double LD; typedef pair<int,int> pii; typedef pair<LL,int>…