题目链接:

啊哈哈,点我点我

题意:

有n个幼儿园的孩纸。然后从中找出m对孩子可以让他们看到两方,这样以便他们交流。。

思路:

首先能够考虑把n-1个人已经排成了m-2对。那么仅仅须要把这个最矮的随便插在队伍就能够凑成m对了。第二种情况是先排成m-1对。然后把最矮的那一个放在对首或者队尾。这样就到了状态转移方程。

if(j>=2)

dp[i][j]=dp[i-1][j-2]*(i-2)

if(j>=1)

dp[i][j]=dp[i][j]+dp[i-1][j-1]*2;

然后最開始把1个人2个人的全部状态枚举出来。。

然后对题目中给的范围进行预处理得到全部解。然后直接询问就可以。问题就得到了完美的解决。

题目:

Queue
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 406   Accepted: 179

Description

Linda is a teacher in ACM kindergarten. She is in charge of n kids. Because the dinning hall is a little bit far away from the classroom, those n kids have to walk in line to the dinning hall every day. When they are walking in line, if and only if two kids
can see each other, they will talk to each other. Two kids can see each other if and only if all kids between them are shorter then both of them, or there are no kids between them. Kids do not only look forward, they may look back and talk to kids behind them.
Linda don’t want them to talk too much (for it’s not safe), but she also don’t want them to be too quiet(for it’s boring), so Linda decides that she must form a line in which there are exactly m pairs of kids who can see each other. Linda wants to know, in
how many different ways can she form such a line. Can you help her?

Note: All kids are different in height.

Input

Input consists of multiple test cases. Each test case is one line containing two integers. The first integer is n, and the second one is m. (0 < n <= 80, 0 <= m <= 10000). 

Input ends by a line containing two zeros.

Output

For each test case, output one line containing the reminder of the number of ways divided by 9937. 

Sample Input

1 0
2 0
3 2
0 0

Sample Output

1
0
4

Source

代码为:

#include<cstdio>
#include<cstring>
#include<iostream>
#define mod 9937
using namespace std; int dp[80+10][10000+10]; void init()
{
memset(dp,0,sizeof(dp));
dp[1][0]=1;
dp[2][1]=2;
for(int i=3;i<=81;i++)
for(int j=1;j<=10001;j++)
{
if(j>=2) dp[i][j]=(dp[i-1][j-2]*(i-2))%mod;
if(j>=1) dp[i][j]=(dp[i][j]+dp[i-1][j-1]*2)%mod;
}
} int main()
{
int n,m;
init();
while(~scanf("%d%d",&n,&m))
{
if(n==0&&m==0) return 0;
cout<<dp[n][m]<<endl;
}
return 0;
}

poj3934Queue(dp)的更多相关文章

  1. LightOJ 1033 Generating Palindromes(dp)

    LightOJ 1033  Generating Palindromes(dp) 题目链接:http://acm.hust.edu.cn/vjudge/contest/view.action?cid= ...

  2. lightOJ 1047 Neighbor House (DP)

    lightOJ 1047   Neighbor House (DP) 题目链接:http://acm.hust.edu.cn/vjudge/contest/view.action?cid=87730# ...

  3. UVA11125 - Arrange Some Marbles(dp)

    UVA11125 - Arrange Some Marbles(dp) option=com_onlinejudge&Itemid=8&category=24&page=sho ...

  4. 【POJ 3071】 Football(DP)

    [POJ 3071] Football(DP) Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 4350   Accepted ...

  5. 初探动态规划(DP)

    学习qzz的命名,来写一篇关于动态规划(dp)的入门博客. 动态规划应该算是一个入门oier的坑,动态规划的抽象即神奇之处,让很多萌新 萌比. 写这篇博客的目标,就是想要用一些容易理解的方式,讲解入门 ...

  6. Tour(dp)

    Tour(dp) 给定平面上n(n<=1000)个点的坐标(按照x递增的顺序),各点x坐标不同,且均为正整数.请设计一条路线,从最左边的点出发,走到最右边的点后再返回,要求除了最左点和最右点之外 ...

  7. 2017百度之星资格赛 1003:度度熊与邪恶大魔王(DP)

    .navbar-nav > li.active > a { background-image: none; background-color: #058; } .navbar-invers ...

  8. Leetcode之动态规划(DP)专题-详解983. 最低票价(Minimum Cost For Tickets)

    Leetcode之动态规划(DP)专题-983. 最低票价(Minimum Cost For Tickets) 在一个火车旅行很受欢迎的国度,你提前一年计划了一些火车旅行.在接下来的一年里,你要旅行的 ...

  9. 最长公共子序列长度(dp)

    /// 求两个字符串的最大公共子序列长度,最长公共子序列则并不要求连续,但要求前后顺序(dp) #include <bits/stdc++.h> using namespace std; ...

随机推荐

  1. hdu 5495 LCS

    Problem Description You are given two sequence {a1,a2,...,an} and {b1,b2,...,bn}. Both sequences are ...

  2. C语言结构体的内存对齐问题

    在C语言开发当中会遇到这样的情况: #include <stdio.h> struct test { int a; char b; }; int main(int argc, const ...

  3. Can you solve this equation?(二分)

    Can you solve this equation? Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Ja ...

  4. js 写日期

    <SCRIPT language="JavaScript" type="text/JavaScript">   today = new Date() ...

  5. wamp安装

    下载之后双击文件进行安装选择:I accept the agreement ,点击Next. 一直单击NEXT 安装完成后运行wamp,在桌面右下角即会出现wamp的图标,图标最初是红色的,然后变为橙 ...

  6. sql相关操作

    1.两个不同数据库对应字段相应操作 //操作模版:insert into data2.table2(字段1,字段2,字段) select 字段j,字段k,字段m from data1.table1举例 ...

  7. (转)Java字符串应用之密码加密与验证

    1.通过java.Security.MessageDigest的静态方法getInstance创建具有指定算法名称的信息摘要,参数为算法名,传入”MD5“则表示使用MD5算法    2.Message ...

  8. makefile常用指令和常见变量。

    引用文章A:http://blog.csdn.net/liang13664759/article/details/1771246 文章介绍:非常详细的文章,讲解上都是比较基础的知识. 本文可能会持续更 ...

  9. JavaScript中setTimeout和setInterval的使用

    相同点:这两个方法都可以用来实现在一个固定的时间之后去实现JavaScript代码,两个方法都包含有两个参数,第一个是将要执行的代码字符串,第二是以毫秒为单位的时间间隔,当过了这个时间间隔之后就会执行 ...

  10. primary key与unique的区别

    定义了 UNIQUE 约束的字段中不能包含重复值,可以为一个或多个字段定义 UNIQUE 约束.因此,UNIQUE 即可以在字段级也可以在表级定义, 在UNIQUED 约束的字段上可以包含空值.ORA ...