取代 fin flutter pan with 最大值 ssi ted
Twilight Sparkle was playing Ludo with her friends Rainbow Dash, Apple Jack and Flutter Shy. But she kept losing. Having returned to the castle, Twilight Sparkle became interested in the dice that were used in the game.
The dice has m faces: the first face of the dice contains a dot, the second one contains two dots, and so on, them-th face containsm dots. Twilight Sparkle is sure that when the dice is tossed, each face appears with probability. Also she knows that each toss is independent from others. Help her to calculate the expected maximum number of dots she could get after tossing the dicen times.
Input
A single line contains two integers m andn (1?≤?m,?n?≤?105).
Output
Output a single real number corresponding to the expected maximum. The answer will be considered correct if its relative or absolute error doesn't exceed10??-?4.
Sample Input
Input
- 6 1
Output
- 3.500000000000
Input
- 6 3
Output
- 4.958333333333
Input
- 2 2
Output
- 1.750000000000
Hint
Consider the third test example. If you've made two tosses:
The probability of each outcome is 0.25, that is expectation equals to:
题目大意:
求出最大点数的期望(这两个字被平均取代后瞬间没有高尚感 - -)。
思路:
若有 5 个的 6 面骰子出现最大点为 1 的次数为 1^5, 出现最大点为 2 的的次数为 2^5-1^5, 以此来推出最大点出现的次数为 i^5-(i-1)^5;
求出最大值得期望,由演示样例能够得到 p = ((1^n-0^n)*1 + (2^n-1^n)*2 。
。
。 +(m^n - (m-1)^n)*m ) / m^n ; 将 m^n 带入 终于得到 p = m - ((m-1)/m)^n + ((m-2)/m)^n。
。。+(1/m)^n ;
- #include<iostream>
- #include<cstring>
- #include<cstdio>
- #include<string>
- #include<cmath>
- #include<map>
- #include<queue>
- #include<algorithm>
- #define LL long long
- int a[1010];
- using namespace std;
- int main()
- {
- LL s;
- int i,j,n,m,k,inr,x;
- double sum;
- while(scanf("%d%d",&m,&n)!=EOF)
- {
- sum=0;
- for(i=1;i<=m;i++)
- {
- sum+=i*(pow(i*1.0/m,n)-pow((i-1)*1.0/m,n));<span id="transmark"></span>
- }
- printf("%.12lf\n",sum);
- }
- }
CodeForces 453A
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