Problem Description The "Harry Potter and the Goblet of Fire" will be on show in the next few days. As a crazy fan of Harry Potter, you will go to the cinema and have the first sight, won’t you? Suppose the cinema only has one ticket-office and…

题目链接:http://acm.hdu.edu.cn/showproblem.php? pid=1133 [题意] 电影票50块一张 有m个人手里正好有50块,n个人手里正好有100块,售票厅開始没有钱.问,有多少种排队的方式,能够让每一个人都买上票. (假设售票厅没有50块零钱,则持有100块的人买不了票) [分析] 显然.当m<n的时候,有0种排列方式. 当m>=n的时候: 用0.代表手里仅仅有50块的人,1,代表手里仅仅有100块的. 则0110100 这样的情况不能满足条件(到第三个人…

  题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1133   题意:排队买50块一张的票,初始票台没有零钱可找,有m个人持有50元,n人持有100元,每人编号各不相同.问有多少种排队方案? 题解: 当 m<n时,肯定方案数是0. 当m>=n时,将队伍看成一个栈,持有50的人用0表示,持有100的人用1表示. 对于n+m个数我们能有的总方案数有C(n+m,n)种. 不符合的方案数:(以下是百度百科的解释) 考虑一个含n个1.n个0的2n位二进制数…

题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=1133 题目大意: 有m+n个人去买电影票,每张电影票50元,  m个人是只有50元一张的,  n个人是只有100元一张的, 电影院自己本身是没有零钱的. 那么要收到100元的钱必须找人家50, 那么再次之前就必须 收到一个50元的, 问你有多少种不同的排列方式. (注意: 这里每个人都看成了不同的元素) 题目分析: 我们要是能找人家钱首先必须要有 m >= n 我们dp[m][n] 再加一个人 只…

传送门 [http://acm.hdu.edu.cn/showproblem.php?pid=1133] 题目描述和分析 代码 #include<iostream> #include<string.h> using namespace std; void Multiply(int a[],int z)//大数a[]和小数z相乘,结果存储在a[]中 { int maxn = 2000; int c = 0; for(int j=maxn-1;j>=0;j--)//用z乘以a[]…

题意: 演唱会门票售票处,那里最开始没有零钱.每一张门票是50元,人们只会拿着100元和50元去买票,有n个人是拿着50元买票,m个人拿着100元去买票. n+m个人按照某个顺序按序买票,如果一个人拿着100元买票,而你没有零钱去找给他,那么买票结束. 题目问你,这n+m个人按照某个顺序按序买票,中间买票没有暂停的排队方式有多少种 题解: 我们设dp[i][j]表示一共有i个人,其中有j个人拿着50元买票的有效排队方式 说一下转移方程: 如果第i个人准备在前i-1个人的排队方式基础上拿着50元去…

Buy the Ticket Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 5566    Accepted Submission(s): 2326 Problem Description The "Harry Potter and the Goblet of Fire" will be on show in the nex…

Train Problem II Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Problem Description As we all know the Train Problem I, the boss of the Ignatius Train Station want to know if all the trains come in strict-increasi…

HDU 4828 Grids 思路:能够转化为卡特兰数,先把前n个人标为0.后n个人标为1.然后去全排列,全排列的数列.假设每一个1的前面相应的0大于等于1,那么就是满足的序列,假设把0看成入栈,1看成出栈.那么就等价于n个元素入栈出栈,求符合条件的出栈序列,这个就是卡特兰数了. 然后去递推一下解,过程中须要求逆元去计算 代码: #include <stdio.h> #include <string.h> const int N = 1000005; const long long…

How Many Trees? Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 3382    Accepted Submission(s): 1960 Problem Description A binary search tree is a binary tree with root k such that any node v re…

HDU 4828 Grids 思路:能够转化为卡特兰数,先把前n个人标为0,后n个人标为1.然后去全排列,全排列的数列,假设每一个1的前面相应的0大于等于1,那么就是满足的序列.假设把0看成入栈,1看成出栈.那么就等价于n个元素入栈出栈,求符合条件的出栈序列,这个就是卡特兰数了.然后去递推一下解,过程中须要求逆元去计算 代码: #include <stdio.h> #include <string.h> const int N = 1000005; const long long…

题目链接 分析:打表以后就能发现时卡特兰数, 但是有除法取余. f[i] = f[i-1]*(4*i - 2)/(i+1); 看了一下网上的题解,照着题解写了下面的代码,不过还是不明白,为什么用扩展gcd, 不是用逆元吗.. 网上还有别人的解释,没看懂,贴一下: (a / b) % m = ( a % (m*b)) / b 笔者注:鉴于ACM题目特别喜欢M=1000000007,为质数: 当gcd(b,m) = 1, 有性质: (a/b)%m = (a*b^-1)%m, 其中b^-1是b模m的逆…

题意是求一列连续升序的数经过一个栈之后能变成的不同顺序的数目. 开始时依然摸不着头脑,借鉴了别人的博客之后,才知道这是卡特兰数,卡特兰数的计算公式是:a( n )  =  ( ( 4*n-2 ) / ( n+1 ) * a( n-1 ) ): 用一个二维数组,a[ i ][ 0 ] 表示第 i 个卡特兰数的位数,a[ i ][ j ] ( j != 0) 中存第 i 个卡特兰数从低位到高位的第 j 个数,也就是说数是倒过来存的,输出时要倒着输出. 代码如下: #include<bits/stdc…

Problem Description As we all know the Train Problem I, the boss of the Ignatius Train Station want to know if all the trains come in strict-increasing order, how many orders that all the trains can get out of the railway.   Input The input contains…

Train Problem II Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 5372    Accepted Submission(s): 2911 Problem Description As we all know the Train Problem I, the boss of the Ignatius Train Stati…

Problem Description This is a small but ancient game. You are supposed to write down the numbers 1, 2, 3, ... , 2n - 1, 2n consecutively in clockwise order on the ground to form a circle, and then, to draw some straight line segments to connect them…

一个A和两个B一共可以组成三种字符串:"ABB","BAB","BBA". 给定若干字母和它们相应的个数,计算一共可以组成多少个不同的字符串.  Input每组测试数据分两行,第一行为n(1<=n<=26),表示不同字母的个数,第二行为n个数A1,A2,...,An(1<=Ai<=12),表示每种字母的个数.测试数据以n=0为结束. Output对于每一组测试数据,输出一个m,表示一共有多少种字符串. Sample Inp…

传送门 FXTZ II Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 530    Accepted Submission(s): 280 Problem Description Cirno is playing a fighting game called "FXTZ" with Sanae. Sanae is a ChuS…

Buy the Ticket Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 3614    Accepted Submission(s): 1522 Problem Description The "Harry Potter and the Goblet of Fire" will be on show in the next…

Buy the Ticket Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 4726    Accepted Submission(s): 1993 Problem Description The "Harry Potter and the Goblet of Fire" will be on show in the nex…

Buy the Ticket Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 1886 Accepted Submission(s): 832   Problem Description The \\\\\\\"Harry Potter and the Goblet of Fire\\\\\\\" will be on show i…

Buy the Ticket Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 8838    Accepted Submission(s): 3684 Problem Description The "Harry Potter and the Goblet of Fire" will be on show in the next…

Buy the Ticket Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 5651    Accepted Submission(s): 2357 Problem Description The "Harry Potter and the Goblet of Fire" will be on show in the nex…

题目链接:https://vjudge.net/problem/HDU-1133 Buy the Ticket Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 7427    Accepted Submission(s): 3105 Problem Description The "Harry Potter and the Goblet…

Buy the Ticket Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 4185    Accepted Submission(s): 1759 Problem Description The "Harry Potter and the Goblet of Fire" will be on show in the next…

Buy the TicketTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6517 Accepted Submission(s): 2720 Problem DescriptionThe "Harry Potter and the Goblet of Fire" will be on show in the next few day…

Problem Description There are many students in PHT School. One day, the headmaster whose name is PigHeader wanted all students stand in a line. He prescribed that girl can not be in single. In other words, either no girl in the queue or more than one…

Buy the Ticket Problem Description The "Harry Potter and the Goblet of Fire" will be on show in the next few days. As a crazy fan of Harry Potter, you will go to the cinema and have the first sight, won’t you? Suppose the cinema only has one tic…

Problem Description The "Harry Potter and the Goblet of Fire" will be on show in the next few days. As a crazy fan of Harry Potter, you will go to the cinema and have the first sight, won't you? Suppose the cinema only has one ticket-office and…

设50元的人为+1 100元的人为-1 满足前随意k个人的和大于等于0 卡特兰数 C(n+m, m)-C(n+m, m+1)*n!*m! import java.math.*; import java.util.*; public class Main { /** * @param args */ public static void main(String[] args) { Scanner sc = new Scanner(System.in); int cas = 1; while(tru…