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…

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…

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 Ticket Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 7152    Accepted Submission(s): 2998 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): 8838    Accepted Submission(s): 3684 Problem Description The "Harry Potter and the Goblet of Fire" will be on show in the next…

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…

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…

FXTZ II Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 498    Accepted Submission(s): 266 Problem Description Cirno is playing a fighting game called "FXTZ" with Sanae.  Sanae is a ChuSho…

Sit sit sit 问题描述 在一个XX大学中有NN张椅子排成一排,椅子上都没有人,每张椅子都有颜色,分别为蓝色或者红色. 接下来依次来了NN个学生,标号依次为1,2,3,...,N. 对于每个学生,他会找一张还没有人坐的椅子坐下来.但是如果这张椅子满足以下三个条件他就不会去坐. 1. 这张椅子左右两边都有相邻的椅子 2. 这张椅子左右两边相邻的椅子都不是空的,也就是有人坐下了 3. 这张椅子左右两边相邻的椅子的颜色不同 如果当前的学生找不到椅子坐下,那他就会走掉. 对于当前的某个学生,他可…

---恢复内容开始--- Hearthstone Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1102    Accepted Submission(s): 544 Problem Description Hearthstone is an online collectible card game from Blizzard Ente…

题目链接 Problem Description Galen Marek, codenamed Starkiller, was a male Human apprentice of the Sith Lord Darth Vader. A powerful Force-user who lived during the era of the Galactic Empire, Marek originated from the Wookiee home planet of Kashyyyk as…

#include<stdio.h> #include<algorithm> using namespace std; void cal(int n,int m) { ; m=min(m,n-m); int j=m; ;m--,i--) { ans*=i; } ;i<=j;i++) { ans/=i; } printf("%I64d\n",ans); } int main() { int _case; int n,m; scanf("%d"…

题意:... 析:每次都是有三种放法,1,2,3,根柱子,所以就是3^n次方. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring&…

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…

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 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…

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…

题目链接:Buy a Ticket 题意: 给出n个点m条边,每个点每条边都有各自的权值,对于每个点i,求一个任意j,使得2×d[i][j] + a[j]最小. 题解: 这题其实就是要我们求任意两点的最短路,但是从点的个数上就知道这题不可以用floyd算法,其实多元最短路可以用dijkstra算.@.@!把所有的点的权值和点放到结构体里面,放入优先队列,其实这样就能保证每次拓展到的点就是这个点的最短路(因为是优先队列,保证拓展到的点这时候的值是最小的),其实就是这个点想通就很简单. #inclu…

D. Buy a Ticket time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Musicians of a popular band "Flayer" have announced that they are going to "make their exit" with a world tou…

题目链接  Buy a Ticket 题意   给定一个无向图.对于每个$i$ $\in$ $[1, n]$, 求$min\left\{2d(i,j) + a_{j}\right\}$ 建立超级源点$n+1$, 对于每一条无向边$(x, y, z)$,$x$向$y$连一条长度为$2z$的边,反之亦然. 对于每个$a_{i}$, 从$i$到$n+1$连一条长度为$a_{i}$的边,反之亦然. 然后跑一边最短路即可. #include <bits/stdc++.h> using namespace…

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 a Ticket 题意要求:求出每个城市看演出的最小费用, 注意的一点就是车票要来回的. 题解:dijkstra 生成优先队列的时候直接将在本地城市看演出的费用放入队列里, 然后直接跑就好了,  dis数组存的是, 当前情况下的最小花费是多少. 代码: #include<iostream> #include<cstring> #include<string> #include<queue> #include<vector> #includ…

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…

Buy A Ticket 题目大意 每个点有一个点权,每个边有一个边权,求对于每个点u的\(min(2*d(u,v)+val[v])\)(v可以等于u) solution 想到了之前的虚点,方便统计终点的权值,将所有点和虚点建边,边权不变,这样只需要求虚点到其他点的最短路即可,就将多源最短路问题转换成了单源最短路 #include <cstdio> #include <cstring> #include <queue> using namespace std; stru…

传送门: http://acm.hdu.edu.cn/showproblem.php?pid=1260 Tickets Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 7318    Accepted Submission(s): 3719 Problem Description Jesus, what a great movie! Th…

Tickets Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 711    Accepted Submission(s): 354 Problem Description Jesus, what a great movie! Thousands of people are rushing to the cinema. However,…

Tickets Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 7925    Accepted Submission(s): 4032 Problem Description Jesus, what a great movie! Thousands of people are rushing to the cinema. Howeve…

HDU 1003    相关链接   HDU 1231题解 题目大意:给定序列个数n及n个数,求该序列的最大连续子序列的和,要求输出最大连续子序列的和以及子序列的首位位置 解题思路:经典DP,可以定义dp[i]表示以a[i]为结尾的子序列的和的最大值,因而最大连续子序列及为dp数组中的最大值.   状态转移方程:dp[1] = a[1]; //以a[1]为结尾的子序列只有a[1]:  i >= 2时, dp[i] = max( dp[i-1]+a[i],  a[i] ); dp[i-1]+a[i…