The King’s Ups and Downs

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HDU 4489 The King's Ups and Downs

HDU 4489 The King's Ups and Downs 思路: 状态:dp[i]表示i个数的方案数. 转移方程:dp[n]=∑dp[j-1]/2*dp[n-j]/2*C(n-1,j-1). 代码: #include<bits/stdc++.h> using namespace std; #define ll long long #define pb push_back #define mem(a,b) memset(a,b,sizeof(a)) ; ll dp[N]; ll C(l…

HDU 4489 The King’s Ups and Downs dp

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=4489 The King's Ups and Downs Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) 问题描述 The king has guards of all different heights. Rather than line them up in increasing o…

hdu 4489 The King’s Ups and Downs(基础dp)

The King’s Ups and Downs Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 582 Accepted Submission(s): 409 Problem Description The king has guards of all different heights. Rather than line the…

UVALive 6177 The King's Ups and Downs

The King's Ups and Downs Time Limit: 3000ms Memory Limit: 131072KB This problem will be judged on UVALive. Original ID: 617764-bit integer IO format: %lld Java class name: Main The king has guards of all different heights. Rather than line the…

HDU 4489 The King’s Ups and Downs （DP+数学计数）

题意:给你n个身高高低不同的士兵.问你把他们按照波浪状排列(高低高或低高低)有多少方法数. 析:这是一个DP题是很明显的,因为你暴力的话,一定会超时,应该在第15个时,就过不去了,所以这是一个DP计数问题. 那么我们应该怎么想呢,我们先假设前 i-1 个已经放好了,然后第 i 个一定是最高的,所以,他一定要在前面找一个低后面放上他,肯定不能放在高的后面, 那么状态就有的表示了,d[i][0]表示是以低结尾,d[i][1]是以高结尾,我们假设放第 i 个士兵时,前面有 j 个,那么后面就有 i -…

HDU 4489 The King’s Ups and Downs

http://acm.hdu.edu.cn/showproblem.php?pid=4489 题意:有n个身高不同的人,计算高低或低高交错排列的方法数. 思路:可以按照身高顺序依次插进去. d[i][0]表示i个人以高低结尾的方法数,d[i][1]表示i个人以低高开头的方法数. 将第i个人插入时,当它左边为j个人的时候,右边就是i-1-j,并且左边必须要以高低结尾,右边必须以低高开头.也就是d[i-1][0]*d[i-1][1].当然了,后面还得再乘c(i-1,j),表示选j个人的方法数. #i…