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题目：时间限制:1000ms单点时限:1000ms内存限制:256MB描述给定一个字符串，判断其是否为合法的正则表达式。 一个正则表达式定义为： 1：0是正则表达式，1也是正则表达式。 2：P和Q都是正则表达式，则PQ是正则表达式。 3：P是正则表达式，则(P)是正则表达式 4：P是正则表达式，则P*也是正则表达式

问题描述：DescriptionConsider m natural numbers n1, n2, ..., nm with the property n1 >= n2 >= ... >= nm > 0. We define a Young table as an arrangement in a table of n1 + n2 + ... + nm natural

问题描述:Skyscraper FloorsTime Limit: 10000MS Memory Limit: 10000KTotal Submissions: 810 Accepted: 175Description What a great idea it is to build skyscrapers! Using not too large area of la

问题描述：Game PredictionTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 9680 Accepted: 4626DescriptionSuppose there are M people, including you, playing a special card game. At the

DescriptionOne way of achieving a smooth and economical drive to work is to `catch' every traffic light, that is have every signal change to green as you approach it. One day you notice as you com

问题描述：Dreisam EquationsTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 2060 Accepted: 418DescriptionDuring excavations in the Dreisamwuste, a desert on some far away and probabl

You Who?Time Limit: 1000MSMemory Limit: 10000KTotal Submissions: 702Accepted: 249Description On the first day of first grade at Friendly Elementrary School, it is customary for each student

问题描述：Complicated ExpressionsTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 1503 Accepted: 485Description The most important activity of ACM is the GSM network. As the mobile pho

问题描述：LCM (Least Common Multiple) of a set of integers is defined as the minimum number, which is a multiple of all integers of that set. It is interesting to note that any positive integer can be

问题描述：A computer programmer lives in a street with houses numbered consecutively (from 1) down one side of the street. Every evening she walks her dog by leaving her house and randomly turning left

UVA 571 Jugs问题描述：In the movie ``Die Hard 3", Bruce Willis and Samuel L. Jackson were confronted with the following puzzle. They were given a 3-gallon jug and a 5-gallon jug and were asked to fil

问题描述：The New VillaTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 1354 Accepted: 443Description Mr. Black recently bought a villa in the countryside. Only one thing bothers him: 

#include#include#includeusing namespace std;ifstream fin("C:\\data12_1.in");struct rect{rect(int a=0,int b=0):length(a),width(b){}int length;int width;};rect arr[100]

写了一个类库，有两个版本：版本一:#include#include#includeusing namespace std;template class Schedule{public: Schedule():cnt(0){} ~Schedule(){} type GetMax(); type ExtractMax(); void Insert(type); bo

本算法是对《算法导论》相关章节伪代码的实现：这是第一次写的：#include#include#include#includeusing namespace std;#define MAX 32767 ifstream fin("data.in");struct node{ node(int x1,int y1,int len1):x(x1),y(y1),len(

问题描述：DescriptionA small boy named Tommy has some toy centipedes that are a series of 1 centimeter segments. Tommy assembles his centipedes to any length he likes and places them on a 30x30 cen

本算法是对＜算法导论＞相关章节伪代码的实现：先贴代码：#include#include#includeusing namespace std;class FibonacciHeap;class node{ friend class FibonacciHeap; node(int v):key(v),height(0),parent(NULL),left(this),righ

本算法是对＜算法导论＞相关章节的实现：下面贴代码：#include#includeusing namespace std;struct node{ node(int d):data(d),degree(0),parent(NULL),child(NULL),sibling(NULL){} int data; int degree; node* parent; node*

算法导论第２３章第２节#include#includeusing namespace std;struct node{ node(int i,int j,int l):x(i),y(j),len(l){} int x; int y; int len;};int N,M;int u,v,l;node* length[1000];int p[1000];int cnt

本题来源：《算法竞赛入门》第11章11.11.2本题给的解法：核心代码：int build_tree(string exp,int l,int r){ cout<<"l="<<l<<"\tr="<<r<<endl; int c1,c2,p,u; c1=-1; c2=-1; p=0; if(r-l<=1) { u=++cnt; chl[u]=chr[u]=0;

此算法为＜算法导论＞中第２４章第２小节关于single-source shortest paths in directed graphs原理的实现．原著P592#includeusing namespace std;#define MAX 32767int N,M;int u,v,l,pos;int w[1000][1000],d[1000];int order[1000]

本算法是对＜算法导论＞第２５章第一节的实现有三个版本：第一个，普通版本：//SLOW-ALL-PAIRS-SHORTEST-PATHS#includeusing namespace std;#define MAX 32767int N,M;int u,v,l;int W[1000][1000],Q[1000][1000],tmp[1000][1000];void p

本算法是对算法导论上相关章节伪代码的实现:#include#include#includeusing namespace std;#define MAX 32767ifstream fin("data.in");int N,M;int u,v,l;int D[1000][1000],p[1000][1000],tmpD[1000][1000],tmpP[1000][1000

Problem BThe Twin TowersInput: standard inputOutput: standard output Once upon a time, in an ancient Empire, there were two towers of dissimilar shapes in two different cities. The towers we

问题描述：An important topic nowadays in computer science is cryptography. Some people even think that cryptography is the only important field in computer science, and that life would not matter at all

问题描述：Problem APlay with Floor and CeilInput: standard inputOutput: standard outputTime Limit: 1 second TheoremFor any two integers x and k there exists two more integers p and q such

#include#include#includeusing namespace std;struct node{       node():frequence(0),ID(' '),left(NULL),right(NULL){}       int frequence;       char ID;       node* left;       

#include#include#includeusing namespace std;ifstream fin("C:\\data11.in");ofstream fout("C:\\data11.out");int n;int arr[1000000];int equalsum[100];int numkind,totalnum;

#include#include#includeusing namespace std;ifstream fin("C:\\data14.in");int n,S;int V[100];int scheme[10000];int value[100];int prev[10000];int trans[10000];void So

//此版本和wode#include#include#includeusing namespace std;ifstream fin("C:\\data14.in");int S;int V[1000];int n;int total[1000];int scheme[10000];void Init(){int valu

#include#include#includeusing namespace std;ifstream fin("C:\\data15.in");int totalVolumn;int weight[100],volumn[100];int n;int scheme[10000];int num[100];void Init()

//这个wom#include#include#includeusing namespace std;ifstream fin("C:\\data12.in");int data[32767];int prev[32676];int maxsum[32767];int n;void Init(){fin>>n;in

#include#include#includeusing namespace std;ifstream fin("C:\\data9.in");string s,t;bool IsSubstring(){     int i,j;     i=0;     for(j=0;j     {          if(t[j]==

#include#include#includeusing namespace std;ifstream fin("C:\\data10.in");struct node{       node(int beg=0,int end=0):startpos(beg),endpos(end){}       node& operator=(const n

//这是按照算法导论给出解法写出的算法#include#include#includeusing namespace std;ifstream fin("C:\\data17.in");char ID[100];int length[100],width[100];int n;int multinum[100][100];int s[100][100];#define M

//此代码为中级版，尚待修改#include#include#include#includeusing namespace std;ifstream fin("C:\\data4.in");ofstream fout("C:\\data4.out");int COUNT=0;char node[8]

//同样k#include#include#includeusing namespace std;ifstream fin("C:\\data13.in");struct rect{rect(int a=0,int b=0):length(a),width(b){}int length;int width;};rect

//此为按贪心算法得出的一种su

#include#include#include#include#includeusing namespace std;#define _DEBUG#ifdef _DEBUGifstream fin("C:\\data5.in");ofstream fout("C:\\data5.out");#

#include#include#includeusing namespace std;ifstream fin("C:\\data6.in");int arr[32767];int cnt;int m;int sum;int cn=0;void Init(){     int x;     cnt=0;     su