HDOJ 2407 Knots(递推规律题)
来源:互联网 发布:如来藏 知乎 编辑:程序博客网 时间:2024/05/20 17:06
Knots
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 219 Accepted Submission(s): 149
Problem Description
An even number N of strands are stuck through a wall. On one side of the wall, a girl ties N/2 knots between disjoint pairs of strands. On the other side of the wall, the girl's groom-to-be also ties N/2 knots between disjoint pairs of strands. You are to find the probability that the knotted strands form one big loop (in which case the couple will be allowed to marry).
For example, suppose that N = 4 and you number the strands 1, 2, 3, 4. Also suppose that the girl has created the following pairs of strands by tying knots: {(1, 4), (2,3)}. Then the groom-to-be has two choices for tying the knots on his side: {(1,2), {3,4)} or {(1,3), (2,4)}.
For example, suppose that N = 4 and you number the strands 1, 2, 3, 4. Also suppose that the girl has created the following pairs of strands by tying knots: {(1, 4), (2,3)}. Then the groom-to-be has two choices for tying the knots on his side: {(1,2), {3,4)} or {(1,3), (2,4)}.
Input
The input file consists of one or more lines. Each line of the input file contains a positive even integer, less than or equal to 100. This integer represents the number of strands in the wall.
Output
For each line of input, the program will produce exactly one line of output: the probability that the knotted strands form one big loop, given the number of strands on the corresponding line of input. Print the probability to 5 decimal places.
Sample Input
420
Sample Output
0.666670.28377
水题,画图找出规律即可。递推公式为a[n]=a[n-1]*((n-2)/(n-1)),且首项a[2]=1.0。
具体代码如下:
<span style="font-size:18px;">#include<cstdio>#include<iostream>using namespace std;int main(){int n,i;double a[105]; a[2]=1.0/1;for(i=4;i<105;i=i+2) a[i]=a[i-2]*(i-2)/(i-1);while(scanf("%d",&n)!=EOF)printf("%.5lf\n",a[n]);return 0;} </span>
0 0
- HDOJ 2407 Knots(递推规律题)
- hdoj--2407--Knots(规律题)
- Knots 2407 (规律题)
- ♥HDOJ 2407-Knots【大爷的规律题】
- hdoj 2407 Knots
- HDOJ-2050(递推,找规律)(折线分割平面)
- 杭电OJ(HDOJ)2041题:超级阶梯(规律递推,Fibonacci)
- 杭电OJ(HDOJ)1005题:Number Sequence(规律递推)
- HDOJ 题目1165 Eddy's research II(递推,找规律)
- 【NOIP 模拟题】中位数(规律+递推)
- 【NOIP 模拟题】求和 (打表找规律+递推)
- URAL 1225-Flags(规律递推)
- hdu6129Just do it(递推规律)
- 找规律 递推
- 递归 递推 规律
- Hdu 2407 Knots【概率】
- hdoj--1005--Number Sequence(规律题)
- HDU2058 找规律+递推
- Linux2.6.32内核笔记(5)在应用程序中移植使用内核链表
- fatal error C1004: unexpected end of file found
- C# get post区别
- springMVC注解的简单使用
- (标记)js中的事件委托
- HDOJ 2407 Knots(递推规律题)
- iOS基础知识:Objective-C 之 谓词
- linux已安装Oracle未安装实例
- golang 子类调用父类函数
- UITableView中自定义Cell,自适应高度
- 欢迎使用CSDN-markdown编辑器
- mysql 数据表读锁机制详解
- 关于JS中prototype的理解
- JQuery中的bind、delegate、on、live方法的区别及简单介绍