题意 有n个循环 给出x a b c xi+1=(a*x+b)%c 要求是从这些循环中各取一个数 使加和最大并且给出一个m 满足sum%m!=0 n的范围是4次方 c的范围是3次方 训练赛的时候看了一眼就觉得好难 稍微一处理就要超时.. 天气好热 又wa了水题两次 十分不想做题QAQ 看着有点难就懒得想 掏出手机开始玩 坑了蕾姐一把 QAQ 以后再也不这样了 然而蕾姐还是机智的想出了几近正解的办法QAQ 在这个循环中 每两个紧挨着的数的间隔必定是一样的 这些数的值在0~C之间 是1000 我们对…

UVALive - 4108 SKYLINE Time Limit: 3000MS     64bit IO Format: %lld & %llu Submit Status uDebug Description   The skyline of Singapore as viewed from the Marina Promenade (shown on the left) is one of the iconic scenes of Singapore. Country X would a…

UVALive - 3942 Remember the Word A potentiometer, or potmeter for short, is an electronic device with a variable electric resistance. It has two terminals and some kind of control mechanism (often a dial, a wheel or a slide) with which the resistance…

UVALive - 3942 Remember the Word Neal is very curious about combinatorial problems, and now here comes a problem about words. Know- ing that Ray has a photographic memory and this may not trouble him, Neal gives it to Jiejie. Since Jiejie can’t remem…

题目传送门 /* 题意:本来有n个雕塑,等间距的分布在圆周上,现在多了m个雕塑,问一共要移动多少距离: 思维题:认为一个雕塑不动,视为坐标0,其他点向最近的点移动,四舍五入判断,比例最后乘会10000即为距离: 详细解释:http://www.cnblogs.com/zywscq/p/4268556.html */ #include <cstdio> #include <iostream> #include <algorithm> #include <cmath&…

题目链接:https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=4156 题目拷贝难度大我就不复制了. 题目大意:维护一个字符串,要求支持插入.删除操作,还有输出第 i 次操作后的某个子串.强制在线. 思路1:使用可持久化treap可破,详细可见CLJ的<可持久化数据结构的研究>. 思路2:rope大法好,详见:http…

Permutation Graphs Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu Submit Status Practice UVALive 6508 #include<stdio.h> #include<string.h> ],a[],b[],c[],b1[]; long long num; void merg_sort(int a[],int l,int r) { int…

Boxes Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu Submit Status Practice UVALive 6500 #include<stdio.h> #include<string.h> int main() { int T,m,n; int i,j,s; ][]; scanf("%d",&T); while(T--) { s=; sc…

题目链接:UVALive 6948  Jokewithpermutation 题意:给一串数字序列,没有空格,拆成从1到N的连续数列. dfs. 可以计算出N的值,也可以直接检验当前数组是否合法. #include <stdio.h> #include <iostream> #include <string.h> #define maxn 100 using namespace std; char str[maxn]; int num[maxn]; bool vis[m…

UVAlive 3135 Argus Argus Time Limit: 3000MS   Memory Limit: Unknown   64bit IO Format: %lld & %llu Submit Status Description A data stream is a real-time, continuous, ordered sequence of items. Some examples include sensor data, Internet traffic, fin…

// 状压DP uvalive 6560 // 题意:相邻格子之间可以合并,合并后的格子的值是之前两个格子的乘积,没有合并的为0,求最大价值 // 思路: // dp[i][j]:第i行j状态下的值 // j:0表示不合并,1表示向下合并 // 一开始输入要修改一下,然后滚动数组优化 #include <iostream> #include <algorithm> #include <cstring> #include <cstdio> #include &…

// 二分+最短路 uvalive 3270 Simplified GSM Network(推荐) // 题意:已知B(1≤B≤50)个信号站和C(1≤C≤50)座城市的坐标,坐标的绝对值不大于1000,每个城市使用最近的信号站.给定R(1≤R≤250)条连接城市线路的描述和Q(1≤Q≤10)个查询,求相应两城市间通信时最少需要转换信号站的次数. // 思路:建议先阅读 NOI论文 <<计算几何中的二分思想>> // 直接献上题解吧: // 二分! // l的两端点所属信号站相同:…

UVAlive 3026 Period 题目: Period   Time Limit: 3000MS   Memory Limit: Unknown   64bit IO Format: %lld & %llu Submit Status Description For each prefix of a given string S with N characters (each character has an ASCII code between 97 and 126, inclusive…

UVAlive 3942 Remember the Word 题目: Remember the Word   Time Limit: 3000MS   Memory Limit: Unknown   64bit IO Format: %lld & %llu Submit Status Description Neal is very curious about combinatorial problems, and now here comes a problem about words. Kn…

UVAlive 4329 Ping pong 题目: Ping pong Time Limit: 3000MS   Memory Limit: Unknown   64bit IO Format: %lld & %llu Submit Status Description N(3N20000) ping pong players live along a west-east street(consider the street as a line segment). Each player ha…

UVAlive 3027 Corporative Network 题目:   Corporative Network Time Limit: 3000MS   Memory Limit: 30000K Total Submissions: 3450   Accepted: 1259 Description A very big corporation is developing its corporative network. In the beginning each of the N ent…

UVAlive X-Plosives 思路:    “如果车上存在k个简单化合物,正好包含k种元素,那么他们将组成一个易爆的混合物”  如果将(a,b)看作一条边那么题意就是不能出现环,很容易联想到Kruskal算法中并查集的判环功能(新加入的边必须属于不同的两个集合否则出现环),因此本题可以用并查集实现.模拟装车过程即可. 代码: #include<cstdio> #include<cstring> #define FOR(a,b,c) for(int a=(b);a<(c…

UVAlive 4794 Sharing Chocolate 题目: http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=12055 思路:   设d[S][r][c]表示形如r*c的矩形是否可以划分为S中的子集,10表示可否. 转移方程: d[S][r][c] = d[S0][r0][c] || d[S0][r][c0]  优化:    首先注意到S r c三者知二求一,所以将状态优化为d[S][x]表示有短边x的矩形是否可以…

 UVAlive 3983 Robotruck 题目: Robotruck   Time Limit: 3000MS   Memory Limit: Unknown   64bit IO Format: %lld & %llu Submit Status Description   Problem C - Robotruck Background This problem is about a robotic truck that distributes mail packages to sev…

UVAlive 4670 Dominating Patterns 题目:   Dominating Patterns   Time Limit: 3000MS   Memory Limit: Unknown   64bit IO Format: %lld & %llu Submit Status Description The archaeologists are going to decipher a very mysterious ``language". Now, they kno…

经典的稳定婚姻匹配问题 UVALive - 3989 Ladies' Choice Time Limit: 6000MS Memory Limit: Unknown 64bit IO Format: %lld & %llu [Submit]   [Go Back]   [Status] Description Problem I – LadiesÕ Choice Background Teenagers from the local high school have asked you to h…

题目 一开始有一个双头队列,每次添加一个数(这是数是二的幂,所有数的和不大于\(2^13\)),由你来决定添加到队头还是队尾.如果队列里面相邻的两个数相同,设它们都是\(x\),那么这两个数会合并为\(2x\).问所有数添加完后队列里能否只剩下一个数. 算法 搜索题,但是需要巧妙地记录状态!这种题不可多. 一个显然的是,队列里不会存在相邻的三个数\(a,b,c\),满足\(a>b,c>b\).这样的话,队列肯定是一个倒V字. 记状态\((i,j)\)为添加完前\(i\)个数,\(j\)是倒V左…

UVALive - 3263 That Nice Euler Circuit (几何) ACM 题目地址:  UVALive - 3263 That Nice Euler Circuit 题意:  给出一个点,问连起来后的图形把平面分为几个区域. 分析:  欧拉定理有:设平面图的顶点数.边数.面数分别V,E,F则V+F-E=2  大白的题目,做起来还是非常有技巧的. 代码: /* * Author: illuz <iilluzen[at]gmail.com> * File: LA3263.cp…

这个题目的深搜形式,我也找出来了,dfs(i,j)表示第i个人选到了第j个物品,但是我却无限RE了,原因是我的viod型深搜太过暴力,我当时定义了一个计数器,来记录并限制递归的层数,发现它已经递归到了1500层,加上限制后,WA了……后来学习了网上的方法,使用bool型的深搜,每一次选择都去跟题目中给的限制去比较,看这次选择有没有冲突,如果没有搜下一个,当搜到false的时候,及时停止,节省了时间和空间. 伪代码: if(dfs(next)==true) return true: else re…

树形背包.DP递推的思路很简单.... 但是由于节点有15万个,先不论空间复杂度,这样开dp数组 dp[150000+10][300+10],如果初始化是memset(dp,-1,sizeof dp),则必然超时. 所以需要一个状态数剪枝...即记录这个节点最多组合的数量. UVALive是不限制内存的,所以dp[150000+10][300+10] 能够AC,HDU 4169 限制了内存大小,需要优化空间复杂度. 内存优化之后的代码,HDU上C++能AC,G++依旧MLE. #include<…

POJ 1330 Nearest Common Ancestors / UVALive 2525 Nearest Common Ancestors (最近公共祖先LCA) Description A rooted tree is a well-known data structure in computer science and engineering. An example is shown below: In the figure, each node is labeled with an…

POJ 3342 Party at Hali-Bula / HDU 2412 Party at Hali-Bula / UVAlive 3794 Party at Hali-Bula / UVA 1220 Party at Hali-Bula(树型动态规划) Description Dear Contestant, I'm going to have a party at my villa at Hali-Bula to celebrate my retirement from BCM. I w…

POJ 2296 Map Labeler / ZOJ 2493 Map Labeler / HIT 2369 Map Labeler / UVAlive 2973 Map Labeler(2-sat 二分) Description Map generation is a difficult task in cartography. A vital part of such task is automatic labeling of the cities in a map; where for e…

Problem   UVALive - 3713 - Astronauts Time Limit: 3000 mSec Problem Description Input The input contains several blocks of test cases. Each case begins with a line containing two integers 1 ≤ n ≤ 100000 and 1 ≤ m ≤ 100000. The number n is the number…

Problem   UVALive - 3211 - Now or later Time Limit: 9000 mSec Problem Description Input Output Sample Input 10 44 156 153 182 48 109 160 201 55 186 54 207 55 165 17 58 132 160 87 197 Sample Output 10 题解:2-SAT问题板子题,这个问题主要是理论难度比较大,有了结论之后代码很容易,有专门的论文阐释算…