light oj 1248 - Dice (III)(概率dp)
来源:互联网 发布:淘宝4px递四方速递运费 编辑:程序博客网 时间:2024/04/28 11:44
Given a dice with n sides, you have to find the expected number of times you have to throw that dice to see all its faces at least once. Assume that the dice is fair, that means when you throw the dice, the probability of occurring any face is equal.
For example, for a fair two sided coin, the result is 3. Because when you first throw the coin, you will definitely see a new face. If you throw the coin again, the chance of getting the opposite side is 0.5, and the chance of getting the same side is 0.5. So, the result is
1 + (1 + 0.5 * (1 + 0.5 * ...))
= 2 + 0.5 + 0.52 + 0.53 + ...
= 2 + 1 = 3
Input
Input starts with an integer T (≤ 100), denoting the number of test cases.
Each case starts with a line containing an integer n (1 ≤ n ≤ 105).
Output
For each case, print the case number and the expected number of times you have to throw the dice to see all its faces at least once. Errors less than 10-6 will be ignored.
Sample Input
Output for Sample Input
5
1
2
3
6
100
Case 1: 1
Case 2: 3
Case 3: 5.5
Case 4: 14.7
Case 5: 518.7377517640
一个均匀的骰子有n个面 投色子 要求最后要把骰子的每一面都看到了 求扔骰子次数的期望
扔骰子只会出现两种情况 一种是扔出了已经出现的面 一种就是扔出了新的面
dp[i]表示扔出i面需要次数的期望 则可以得到dp[i]=i/n*dp[i]+(n-i)/n*dp[i+1]+1
得到dp[i]=dp[i+1]+n/(n-i) 其中dp[n]=0 答案为dp[0]
#include <cstdio>#include <iostream>#include <cstring>#include <cmath>#include <algorithm>#include <string.h>#include <string>#include <vector>#include <queue>#define MEM(a,x) memset(a,x,sizeof a)#define eps 1e-8#define MOD 10009#define MAXN 10010#define MAXM 100010#define INF 99999999#define ll __int64#define bug cout<<"here"<<endl#define fread freopen("ceshi.txt","r",stdin)#define fwrite freopen("out.txt","w",stdout)using namespace std;int Read(){ char c = getchar(); while (c < '0' || c > '9') c = getchar(); int x = 0; while (c >= '0' && c <= '9') { x = x * 10 + c - '0'; c = getchar(); } return x;}void Print(int a){ if(a>9) Print(a/10); putchar(a%10+'0');}double dp[100010];int main(){// fread; int tc; scanf("%d",&tc); int cs=1; while(tc--) { int n; scanf("%d",&n); dp[n]=0.0; for(int i=n-1;i>=0;i--) { dp[i]=dp[i+1]+(n*1.0)/((double)(n-i)); } printf("Case %d: %.7lf\n",cs++,dp[0]); } return 0;}
- Light OJ 1248 - Dice (III) 概率DP
- light oj 1248 - Dice (III)(概率dp)
- light oj 1248 - Dice (III)(期望)
- Light oj 1064 Throwing Dice (概率dp)
- 【LightOJ 1248】Dice (III)(概率DP)
- lightOJ 1248 - Dice (III) 概率DP
- light oj 1248 概率dp
- LightOJ1248---Dice (III)(概率dp)
- light oj 1017 - Brush (III) (dp)
- light oj 1145 - Dice (I) DP计数
- Light OJ 1064 Throwing Dice(简单DP)
- Light OJ 1145 Dice (I) (DP)
- Light OJ 1193 Dice (II) (DP)
- Light OJ 1038(概率DP)
- lightOj 1248 Dice (III) 概率公式
- LightOJ1248---Dice (III) (概率问题)
- Light oj 1038 Race to 1 Again(概率dp)
- light oj 1265 - Island of Survival(概率dp)
- Java实现回调函数
- codeforces 551 C GukiZ hates Boxes
- hdu5296(2015多校1)--Annoying problem(lca+一个公式)
- java中如何终止线程
- IAR8.302破解
- light oj 1248 - Dice (III)(概率dp)
- RMQ专题练习
- 队列相关知识
- 【2015/7/22】SqlServer卸载重装全攻略!
- OC基础_数组
- 【转】斐波那契数列取模(大数)分治算法
- 2015.7.22 关于为什么用二级指针作参数
- Linux1.0系统调用列表
- C#计算float类型数据求和不精确问题