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…

  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…

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…

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…

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…

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…

题意:有m个人有一张50元的纸币,n个人有一张100元的纸币.他们要在一个原始存金为0元的售票处买一张50元的票,问一共有几种方案数. 解法:(学习了他人的推导后~) 1.Catalan数的应用7的变形.(推荐阅读:http://www.cnblogs.com/chenhuan001/p/5157133.html).P.S.不知我之前自己推出的公式"C(n,m)*C(2*m,m)/(m+1)*P(n,n)*P(m,m)"是否是正确的. (1)在不考虑m人和n人本身组内的排列时,总方案数…

sequence2 Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=5568 Description Given an integer array bi with a length of n, please tell me how many exactly different increasing subsequences. P.S. A subsequence bai(…

题目连接:http://acm.hdu.edu.cn/showproblem.php?pid=1042 题目大意:求n!, n 的上限是10000. 解题思路:高精度乘法 , 因为数据量比较大, 所以得用到缩进,但因为是乘法,缩进只能在10^5, 不然在乘的过程中就超过int型.然后数值位数大概在8000位以下. #include <iostream> #include <string.h> using namespace std; const int MAXN = 100000;…

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…

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…

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[]…

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…

Sum Time Limit:1000MS     Memory Limit:131072KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 4704 Description   Sample Input 2   Sample Output 2 Hint 1. For N = 2, S(1) = S(2) = 1. 2. The input file consists of multiple test cases. 题意…

Oracle Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others) Problem Description There is once a king and queen, rulers of an unnamed city, who have three daughters of conspicuous beauty. The youngest and most beautifu…

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…

Problem Description Given an integer N(0 ≤ N ≤ 10000), your task is to calculate N!   Input One N in one line, process to the end of file.   Output For each N, output N! in one line.   Sample Input 1 2 3   Sample Output 1 2 6 一看到题目就想到用高精度乘法,用数组模拟小学学过…

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…

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…

题意:有n个药片,每次吃半片,吃2n天,那么有多少种吃法. 分析:如果说吃半片,那么一定要吃过一整片,用 ) 表示吃半片,用 ( 表示吃整片,那么就是求一个正确的括号匹配方案数,即卡特兰数. 卡特兰数可以DP,也有一重循环的地推表达式:Catalan(n) = 2(2n-1)*Catalan(n-1)/(n+1) #include <bits/stdc++.h> using namespace std; ]={,,,}; int main() { //freopen("in.txt&…

题意:每个人有一个DI值,现在有一个小黑屋,这些人的顺序可以利用这个小黑屋调整,调整方式是入栈出栈方式,也就是说,这里的方案是有卡特兰数个方式. 调整后使得 d1*0 + d2*1 + d3*2 + d4*3 ...... 最小. 分析:这个题目竟然会是区间DP. 考虑区间 [ L, R ] ,那么L,可以从任意位置出栈,枚举出栈位置,可以划分为两个部分,也就是说两个子问题,但是如何利用这两个子问题得到 d[L,R], d[L,R] = 前一部分 + D[L]*(i-L) + 后一部分 + 后一…

题意: 给出一个\(n(0 \leq n \leq 10^{12})\),问\(n\)个\(M\)形的折线最多可以把平面分成几部分. 分析: 很容易猜出来这种公式一定的关于\(n\)的一个二次多项式. 不妨设\(f(n)=an^2+bn+c\). 结合样例我们可以列出\(3\)个方程: \(f(0)=1,f(1)=2,f(2)=19\) 解出三个系数\(a,b,c\),然后用高精度做即可. #include <cstdio> #include <cstring> #include…

The Luckiest number Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 980    Accepted Submission(s): 301 Problem Description Chinese people think of '8' as the lucky digit. Bob also likes digit '8…

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…

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…

题意 : 给出四个点,问你第四个点是否在前三个点构成的圆内,若在圆外输出"Accepted",否则输出"Rejected",题目保证前三个点不在一条直线上. 分析 : 简单的计算几何问题,如果能够知道圆心和半径(Radius)以及第四个点和圆心的距离(Distance),我们就能够判断第四个点是否在圆外,例如Distance > Radius则在圆外.三点构圆 的圆心和半径是能够推导出公式的 (参考==> http://blog.csdn.net/dea…

首先,记50的为0,100的为1. 当m=4,n=3时,其中的非法序列有0110010; 从不合法的1后面开始,0->1,1->0,得到序列式0111101 也就是说,非法序列变为了n-1个0,m+1个1. 总的数目=C(m+n,n),非法的=C(m+n,m+1) 符合数目=(C(m+n,n)-C(m+n,m+1))*m!*n!; 化简得:(m+n)!*(m+1-n)/(m+1). Java代码: import java.io.*; import java.math.*; import jav…

//大数继续,额,要吐了. Problem Description Problems involving the computation of exact values of very large magnitude and precision are common. For example, the computation of the national debt is a taxing experience for many computer systems.  This problem r…