Counting-out Rhyme CodeForces - 792B 题意: n 个孩子在玩一个游戏. 孩子们站成一圈,按照顺时针顺序分别被标号为 1 到 n.开始游戏时,第一个孩子成为领导. 游戏进行 k 轮. 在第 i 轮中,领导会从他顺时针方向下一个孩子开始数 ai 个孩子.最后数到的那个孩子出局,再下一个孩子成为新的领导. 举个例子, 现在圈内还剩 [8, 10, 13, 14, 16] 4个孩子,领导编号为 13 , ai = 12.那么出局的孩子为 16 .第 8 个孩子成为下一…

In the city of Ultima Thule job applicants are often offered an IQ test. The test is as follows: the person gets a piece of squared paper with a 4 × 4 square painted on it. Some of the square's cells are painted black and others are painted white. Yo…

题目链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=1368 题目大意:计算从S到T中所有的数,其中0,1,2,3,4,5,6,7,8,9的个数,例如从1024到1032有1024 1025 1026 1027 1028 1029 1030 1031 1032,其中10个0,10个1,7个2,3个3等等 Sample Input 1 1044 497346 5421199 17481496 14031004 503171…

poj 题目分类 按照ac的代码长度分类(主要参考最短代码和自己写的代码) 短代码:0.01K--0.50K:中短代码:0.51K--1.00K:中等代码量:1.01K--2.00K:长代码:2.01K以上. 短:1147.1163.1922.2211.2215.2229.2232.2234.2242.2245.2262.2301.2309.2313.2334.2346.2348.2350.2352.2381.2405.2406: 中短:1014.1281.1618.1928.1961.2054…

本文来自:http://www.cppblog.com/snowshine09/archive/2011/08/02/152272.spx 多版本的POJ分类 流传最广的一种分类: 初期: 一.基本算法: (1)枚举. (poj1753,poj2965) (2)贪心(poj1328,poj2109,poj2586) (3)递归和分治法. (4)递推. (5)构造法.(poj3295) (6)模拟法.(poj1068,poj2632,poj1573,poj2993,poj2996) 二.图算法:…

acm之pku题目分类 对ACM有兴趣的同学们可以看看 DP:  1011   NTA                 简单题  1013   Great Equipment     简单题  1024   Calendar Game       简单题  1027   Human Gene Functions   简单题  1037   Gridland            简单题  1052   Algernon s Noxious Emissions 简单题  1409   Commun…

转载:from: POJ:http://blog.csdn.net/qq_28236309/article/details/47818407 按照ac的代码长度分类(主要参考最短代码和自己写的代码) 短代码:0.01K–0.50K:中短代码:0.51K–1.00K:中等代码量:1.01K–2.00K:长代码:2.01K以上. 短:1147.1163.1922.2211.2215.2229.2232.2234.2242.2245.2262.2301.2309.2313.2334.2346.2348…

B. Counting-out Rhyme time limit per test: 1 second memory limit per test: 256 megabytes input: standard input output: standard output n children are standing in a circle and playing the counting-out game. Children are numbered clockwise from 1 to n.…

ZOJ3944 People Counting ZOJ3939 The Lucky Week 1.PeopleConting 题意:照片上有很多个人,用矩阵里的字符表示.一个人如下: .O. /|\ (.) 占3*3格子,句号“.”为背景.没有两个人完全重合.有的人被挡住了一部分.问照片上有几个人. 题解: 先弄个常量把3*3人形存起来,然后6个部位依次找,比如现在找头,找到一个头,就把这个人删掉(找这个人的各个部位,如果在该部位位置的不是这个人的身体,就不删),删成句号,疯狂找就行了. 代码:…

题目连接:http://codeforces.com/contest/676/problem/B 题意:给你一个N层的杯子堆成的金字塔,倒k个杯子的酒,问倒完后有多少个杯子的酒是满的 题解:由于数据不是很大,直接模拟就行了 #include<cstdio> #include<cstring> #define F(i,a,b) for(int i=a;i<=b;++i) ][],eps=1e-,p;int n,k,ans; void fuck(){ a[][]+=; F(i,,…

Shaass has decided to hunt some birds. There are n horizontal electricity wires aligned parallel to each other. Wires are numbered 1 to nfrom top to bottom. On each wire there are some oskols sitting next to each other. Oskol is the name of a delicio…

考的时候脑子各种短路,用个SAM瞎搞了半天没搞出来,最后中午火急火燎的打了个SPFA才混了点分. 其实这个可以把每个模式串长度为$K-1$的字符串看作一个状态,这个用字符串Hash实现,然后我们发现这实际上可以通过不断转移变成一个DAG,通过topsort即可算出最优解. 我怎么就那么脑残没想到呢.. //rhyme //by Cydiater //2017.2.19 #include <iostream> #include <cstdio> #include <queue&…

Description 羽月最近发现,她发动能力的过程是这样的: 构建一个 V 个点的有向图 G,初始为没有任何边,接下来羽月在脑中构建出一个长度为 E 的边的序列,序列中元素两两不同,然后羽月将这些边依次加入图中,每次加入之后计算当前图的强连通分量个数并记下来,最后得到一个长度为E 的序列,这个序列就是能力的效果. 注意到,可能存在边的序列不同而能力效果相同的情况,所以羽月想请你帮她计算能发动的不同能力个数,答案对 998244353 取模.你需要对于1<=E<=V*(V-1)的所有 E 计…

题目大意:给你一段内存,要你进行如下的三个操作. 1.分配内存  alloc   X ,分配连续一段长度为X的内存. 如果内存不够应该输出NULL,如果内存够就给这段内存标记一个编号. 2.擦除编号为 X的内存,erase X,  如果这段内存不存在那么输出“ILLEGAL_ERASE_ARGUMENT ”,否则什么都不输出. 3.整理内存,把所有的内存块整理到一块.   #include<cstdio> #include<cstring> #include<iostream…

 Devu and Partitioning of the Array time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Devu being a small kid, likes to play a lot, but he only likes to play with arrays. While playing he came…

题目链接 Problem Description Your input is a series of rectangles, one per line. Each rectangle is specified as two points(X,Y) that specify the opposite corners of a rectangle. All coordinates will be integers in the range 0 to 100. For example, the lin…

题目链接:Kilani and the Game 题目原文 Kilani is playing a game with his friends. This game can be represented as a grid of size…

题目链接:Grid game 题目原文 You are given a 4x4 grid. You play a game — there is a sequence of tiles, each of them is either 2x1 or 1x2. Your task is to consequently place all tiles from the given sequence in the grid. When tile is placed, each cell which is…

题目 题目大意自己看题去-- 正解 比赛时在刚第二题,所以根本没有时间思考-- 模型可以转化为从\((x_1,x_2,..,x_n)\)出发到\((1,1)\)的方案数模\(2\). 方案数就用有重复的排列公式:\(\frac{(\sum{x_i})!}{\prod x_i!}\) 考虑它的奇偶性.显然可以将上面的\(2\)因子个数求出来,减去下面的个数,如果为\(0\)则是奇数. 这个东西也就是下面这条式子:\(\sum_{w=2^i} (\lfloor \frac{\sum_{x_i}}{w…

hafy 由于多次交换邮票没有满足所有人的需求,小Z被赶出了集邮部.无处可去的小Z决定加入音乐部,为了让音乐部的人注意到自己的才华,小Z想写一首曲子.为了让自己的曲子更好听,小Z找到了一些好听曲子作为模板.曲谱可以表示成只包含小写字母的字符串,小Z希望自己最终的曲谱中任意一个长度为K的子串都是一个模板的子串.现在小Z想知道自己的曲谱最长可以是多长,如果可以无限长的话请输出INF. forget 对于30%的数据:K=2. 对于70%的数据:每组数据字符串总长不超过1000. 对于100%的数据:…

题意: 输入一个正整数N(N<=2^30),输出从1到N共有多少个数字包括1. AAAAAccepted code: #define HAVE_STRUCT_TIMESPEC #include<bits/stdc++.h> using namespace std; int main(){ ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL); int n; cin>>n; ; ,r=,low_bit=,yushu…

ACM思维题训练集合 You are given two integers n and d. You need to construct a rooted binary tree consisting of n vertices with a root at the vertex 1 and the sum of depths of all vertices equals to d. A tree is a connected graph without cycles. A rooted tre…

Long is a huge fan of CFC (Codeforces Fried Chicken). But the price of CFC is increasing, so he decides to breed the chicken on his own farm. His farm is presented by a rectangle grid with rr rows and cc columns. Some of these cells contain rice, oth…

Bob is an avid fan of the video game "League of Leesins", and today he celebrates as the League of Leesins World Championship comes to an end! The tournament consisted of nn (n≥5n≥5) teams around the world. Before the tournament starts, Bob has…

Description Let S = s1 s2...s2n be a well-formed string of parentheses. S can be encoded in two different ways: q By an integer sequence P = p1 p2...pn where pi is the number of left parentheses before the ith right parenthesis in S (P-sequence). q B…

I Think I Need a Houseboat Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 85149   Accepted: 36857 Description Fred Mapper is considering purchasing some land in Louisiana to build his house on. In the process of investigating the land,…

题目链接: B. Pyramid of Glasses time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Mary has just graduated from one well-known University and is now attending celebration party. Students like to d…

转载请注明出处: http://www.cnblogs.com/fraud/          ——by fraud Always an integer Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu Combinatorics is a branch of mathematics chiefly concerned with counting discrete objects. For instanc…

Roman Roulette Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 286    Accepted Submission(s): 105 Problem Description The historian Flavius Josephus relates how, in the Romano-Jewish conflict of…

题目链接: 啊哈哈,选我选我 思路是: 相当于模拟约瑟夫环,仅仅只是是从顺逆时针同一时候进行的,然后就是顺逆时针走能够编写一个函数,仅仅只是是走的方向的标志变量相反..还有就是为了(pos+flag+n-1)%n+1的妙用... 题目:  The Dole Queue  In a serious attempt to downsize (reduce) the dole queue, The New National Green Labour Rhinoceros Party has decid…