CodeForces 454C Little Pony and Expected Maximum(排列组合)

C. Little Pony and Expected Maximum
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
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, the m-th face contains m 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 dice n times.

Input
A single line contains two integers m and n (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 exceed 10  - 4.

Examples
inputCopy
6 1
output
3.500000000000
inputCopy
6 3
output
4.958333333333
inputCopy
2 2
output
1.750000000000

题意

有一个有m面的骰子,骰子上面的点数为1-m,每一面的概率为$\frac{1}{m}$,请问投掷n次,能得到的最大数的期望

思路

排列的总数是$m^{n}$次,$m^{n}$次里面包括$1\sim m$的所有排列,那么$(m-1)^{n}$就包含了$1\sim m-1$的所有排列,那么自然而然的得到了$m^n-(m-1)^{n}$点数m的所有排列出现的次数,因为题目中求得是最大的数,在所有的m的排列中m就是最大的数所以m的贡献就是$(m^n-(m-1)^{n})m$,同理$m-1$的贡献就是$((m-1)^n-(m-2)^{n})(m-1)$所以我就得到了公式

$$ans=\frac{(m^n-(m-1)^{n})m+\cdots+(2^n-1)2+1}{m^n}$$

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
using namespace std;
int main()
{
    int n,m;
    while(scanf("%d%d",&m,&n)!=EOF)
    {
       double ans=0;
       for(int i=m;i>=1;i--)
       {
           ans+=(pow(1.0*i/m,n)-pow(1.0*(i-1)/m,n))*i;
       }
       printf("%.4lf\n",ans);
    }
    return 0;
}