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http://poj.org/problem?id=2492 Hint Huge input,scanf is recommended. 也是带种类的冰茶几。 由于只分了两类…我们还是可以按照上道题的做法。。 感觉完全是一样的题啊。。 结果一直WA。。。。 最后发现。。。我边读入边判断。。发现同性恋了就直接Break掉了。。。后面改组的数据读到下一组去了233，不WA就日了汪了。。。

http://poj.org/problem?id=1703 种类冰茶几…看到还有一种算是拓展的交加权冰茶几？ 看到有做法是在开一个数组。。。记录是哪一组…. 但是因为只有两组….我们可以分别存… 因为不知道每一个D的两个人分别是哪个组（帮派？） 可以都存一下。 TLE了两次….应该是用了cin的事。。。改成scanf就变WA了。。。 想了下。原来是我对“not sure yet”的判断出现失误。 我开了一个v数组，记录在D下出现的人。 我误以为出现的人的帮派一定是确定的。 实际上并不是。 比如 1,3 5,7 3和7都出现了。但是3和7是一组与否显然还是“not sure yet”

http://codeforces.com/problemset/problem/535/C 题读了好几遍才读懂。 题意是给出一个等差数列，操作严格要求从最左边不为零的连续m个数减去1，最多执行t次后问离最左边最远的位置在哪里。 有两个限制条件…一个是本身的si不能大于t，否则无法吃完。 还有一个是从sl到sr的和不能超过m*t (比赛的时候考虑的不周到。。实际上只有当r-l+1比m大的时候才是m，也就是说要取min(m,l-r+1)) 这题正解应该是二分….直接Lower_bound。。。看到也有人用前缀和搞的。 我是解方程了（貌似是个傻逼做法）…. 可以列出一个关于r的一元二次方程。。。然后求根公式2333 方程是：

http://codeforces.com/problemset/problem/534/C 题意是说一共有N个骰子，第I个筛子一共有di面…现在知道这些骰子的点数之和，问对于每一个骰子不能取得值有多少个。 乍一看有点不明觉厉…稍微再想下，求取值范围即可。

http://codeforces.com/problemset/problem/534/B 题意是说一辆车，每秒内的速度恒定…第I秒到第I+1秒的速度变化不超过D。初始速度为V1，末速度为V2，经过时间t，问最远能走多远。 策略就是尽可能加速…加到某个时间，如果在这个时间不开始减速就回不到V2了。

B. Tavas and SaDDas time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Once again Tavas started eating coffee mix without water! Keione told him that it smells awful, but he didn’t stop doing that. That’s why Keione told his smart friend, SaDDas to punish him! SaDDas took Tavas’ headphones and told him: “If you solve the following problem, I’ll return it to you.” The problem is: You are given a lucky number n. Lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.

A. Exam time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output An exam for n students will take place in a long and narrow room, so the students will sit in a line in some order. The teacher suspects that students with adjacent numbers (i and i + 1) always studied side by side and became friends and if they take an exam sitting next to each other, they will help each other for sure.

http://codeforces.com/problemset/problem/479/D 题意是说有一把尺子，本身有一些刻度，然后需要测量x和y，问最少需要添加多少个刻度，如果需要，这些刻度分别添加在什么位置。 一开始没有看清题目，以为答案最多为4，但是发现，a[1]=0 a[n]=l这两个是确定的，所以答案最多为2，

http://codeforces.com/problemset/problem/525/B 1题意是说一个字符串，进行m次颠倒变换（从a[i]位置到a[l-i+1]位置），问得到的字符串。容易发现，对于越在里边（对称，也就是越靠近中间位置）的字符，调换的次数越多。我们可以把a[i]从小到大排序。然后经过分析发现，把两个相邻的a[i]分为一组，做处理，如果m为奇数，最后还剩下a[m]没有被分组，要单独处理a[m]细节上要注意st数组是从st[0]开始的...好吧的确不方便，适牛也说我了。。数组下标以后还是从0开始吧。。。主要是受高中OI用的pascal的影响。。。那个数组下标随便啊。代码： 2 3 4 5 6 /* *********************************************** 7 Author :111qqz 8 Created Time :2016年02月22日 星期一 23时39分51秒 9 File Name :code/cf/problem/525B.cpp 10 ************************************************ */ 11 12 #include <iostream> 13 #include <algorithm> 14 #include <cstring> 15 #include <cmath> 16 #include <cstdio> 17 18 using namespace std; 19 20 int m,k,len; 21 const int N=2E5+7; 22 int a[N]; 23 char st[N]; 24 25 int main() 26 { 27 cin>>st; 28 scanf("%d",&m); 29 for ( int i = 1 ; i <= m ; i++ ) 30 scanf("%d",&a[i]); 31 sort(a+1,a+m+1); 32 k = 1; 33 len = strlen(st); 34 while (k<=m) 35 { 36 for ( int j = a[k] ; j <= a[k+1]-1 ; j++) 37 swap(st[j-1],st[len-j]); 38 k = k + 2; 39 } 40 if ( m %2==1 ) 41 for ( int i = a[m]; i <= len/2 ; i++ ) 42 swap(st[i-1],st[len-i]); 43 cout<<st<<endl; 44 return 0; 45 }

C - C **Time Limit:**1000MS **Memory Limit:**262144KB 64bit IO Format:%I64d & %I64u Submit Status Description Permutation_p_ is an ordered set of integers _p_1, p_2, …, p__n, consisting of n distinct positive integers not larger than n. We’ll denote as_n the length of permutation _p_1, _p_2, …, p__n. Your task is to find such permutation p of length n, that the group of numbers |_p_1 - _p_2|, |_p_2 - _p_3|, …, |p__n - 1 - p__n| has exactly k distinct elements. Input The single line of the input contains two space-separated positive integers n, k (1 ≤ k < n ≤ 105).

http://codeforces.com/problemset/problem/479/C 1/************************************************ 2Author :111qqz 3Created Time :2016年02月22日 星期一 23时31分10秒 4File Name :code/cf/problem/479C.cpp 5************************************************ */ 6 7#include <iostream> 8#include <algorithm> 9#include <cstring> 10#include <cstdio> 11 12#include <cmath> 13 14using namespace std; 15int n,ans; 16const int N=1E4+5; 17int a[N],b[N]; 18 19struct Q 20{int a,b; 21}q[N]; 22 23bool cmp(Q x, Q y) 24{ 25 if ( x.a<y.a) return true; 26 if ( x.a==y.a &&x.b<y.b ) return true; 27 return false; 28} 29 30int main() 31{ 32 scanf("%d",&n); 33 for ( int i = 1 ; i <= n ; i++ ) 34 scanf("%d %d",&q[i].a,&q[i].b); 35 sort(q+1,q+n+1,cmp); 36 37 ans=q[1].b; 38 for ( int i = 2 ; i <= n; i++ ) 39 { 40 if ( q[i].b>=ans ) 41 ans = q[i].b; 42 else ans = q[i].a; 43 } 44 printf("%d\n",ans); 45 return 0; 46}

FatMouse’s Speed # Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 10172 Accepted Submission(s): 4521 Special Judge Problem Description FatMouse believes that the fatter a mouse is, the faster it runs. To disprove this, you want to take the data on a collection of mice and put as large a subset of this data as possible into a sequence so that the weights are increasing, but the speeds are decreasing.

Tickets # **Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 1408 Accepted Submission(s): 687 ** Problem Description Jesus, what a great movie! Thousands of people are rushing to the cinema. However, this is really a tuff time for Joe who sells the film tickets. He is wandering when could he go back home as early as possible. A good approach, reducing the total time of tickets selling, is let adjacent people buy tickets together. As the restriction of the Ticket Seller Machine, Joe can sell a single ticket or two adjacent tickets at a time. Since you are the great JESUS, you know exactly how much time needed for every person to buy a single ticket or two tickets for him/her. Could you so kind to tell poor Joe at what time could he go back home as early as possible? If so, I guess Joe would full of appreciation for your help.

免费馅饼 # **Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 29065 Accepted Submission(s): 9921 ** Problem Description 都说天上不会掉馅饼，但有一天gameboy正走在回家的小径上，忽然天上掉下大把大把的馅饼。说来gameboy的人品实在是太好了，这馅饼别处都不掉，就掉落在他身旁的10米范围内。馅饼如果掉在了地上当然就不能吃了，所以gameboy马上卸下身上的背包去接。但由于小径两侧都不能站人，所以他只能在小径上接。由于gameboy平时老呆在房间里玩游戏，虽然在游戏中是个身手敏捷的高手，但在现实中运动神经特别迟钝，每秒种只有在移动不超过一米的范围内接住坠落的馅饼。现在给这条小径如图标上坐标：

F - Piggy-Bank **Time Limit:**1000MS **Memory Limit:**32768KB 64bit IO Format:%I64d & %I64u Submit Status Description Before ACM can do anything, a budget must be prepared and the necessary financial support obtained. The main income for this action comes from Irreversibly Bound Money (IBM). The idea behind is simple. Whenever some ACM member has any small money, he takes all the coins and throws them into a piggy-bank. You know that this process is irreversible, the coins cannot be removed without breaking the pig. After a sufficiently long time, there should be enough cash in the piggy-bank to pay everything that needs to be paid.

E - Super Jumping! Jumping! Jumping! **Time Limit:**1000MS **Memory Limit:**32768KB 64bit IO Format:%I64d & %I64u Submit Status Description Nowadays, a kind of chess game called “Super Jumping! Jumping! Jumping!” is very popular in HDU. Maybe you are a good boy, and know little about this game, so I introduce it to you now. The game can be played by two or more than two players. It consists of a chessboard（棋盘）and some chessmen（棋子）, and all chessmen are marked by a positive integer or “start” or “end”. The player starts from start-point and must jumps into end-point finally. In the course of jumping, the player will visit the chessmen in the path, but everyone must jumps from one chessman to another absolutely bigger (you can assume start-point is a minimum and end-point is a maximum.). And all players cannot go backwards. One jumping can go from a chessman to next, also can go across many chessmen, and even you can straightly get to end-point from start-point. Of course you get zero point in this situation. A player is a winner if and only if he can get a bigger score according to his jumping solution. Note that your score comes from the sum of value on the chessmen in you jumping path. Your task is to output the maximum value according to the given chessmen list.

C - Monkey and Banana **Time Limit:**1000MS **Memory Limit:**32768KB 64bit IO Format:%I64d & %I64u Submit Status Description A group of researchers are designing an experiment to test the IQ of a monkey. They will hang a banana at the roof of a building, and at the mean time, provide the monkey with some blocks. If the monkey is clever enough, it shall be able to reach the banana by placing one block on the top another to build a tower and climb up to get its favorite food.

**Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 26534 Accepted Submission(s): 9332 ** Problem Description Nowadays, we all know that Computer College is the biggest department in HDU. But, maybe you don’t know that Computer College had ever been split into Computer College and Software College in 2002. The splitting is absolutely a big event in HDU! At the same time, it is a trouble thing too. All facilities must go halves. First, all facilities are assessed, and two facilities are thought to be same if they have the same value. It is assumed that there is N (0<N<1000) kinds of facilities (different value, different kinds).

I NEED A OFFER! # **Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 18287 Accepted Submission(s): 7320 ** Problem Description Speakless很早就想出国，现在他已经考完了所有需要的考试，准备了所有要准备的材料，于是，便需要去申请学校了。要申请国外的任何大学，你都要交纳一定的申请费用，这可是很惊人的。Speakless没有多少钱，总共只攒了n万美元。他将在m个学校中选择若干的（当然要在他的经济承受范围内）。每个学校都有不同的申请费用a（万美元），并且Speakless估计了他得到这个学校offer的可能性b。不同学校之间是否得到offer不会互相影响。“I NEED A OFFER”，他大叫一声。帮帮这个可怜的人吧，帮助他计算一下，他可以收到至少一份offer的最大概率。（如果Speakless选择了多个学校，得到任意一个学校的offer都可以）。

饭卡 # **Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 14225 Accepted Submission(s): 4945 ** Problem Description 电子科大本部食堂的饭卡有一种很诡异的设计，即在购买之前判断余额。如果购买一个商品之前，卡上的剩余金额大于或等于5元，就一定可以购买成功（即使购买后卡上余额为负），否则无法购买（即使金额足够）。所以大家都希望尽量使卡上的余额最少。 某天，食堂中有n种菜出售，每种菜可购买一次。已知每种菜的价格以及卡上的余额，问最少可使卡上的余额为多少。