E. Little Pony and Expected Maximum（组合期望）
题目描述：
Little Pony and Expected Maximum
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard 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
Input
Copy
6 1
Output
Copy
3.500000000000
Input
Copy
6 3
Output
Copy
4.958333333333
Input
Copy
2 2
Output
Copy
1.750000000000
Note
Consider the third test example. If you've made two tosses:
- You can get 1 in the first toss, and 2 in the second. Maximum equals to 2.
- You can get 1 in the first toss, and 1 in the second. Maximum equals to 1.
- You can get 2 in the first toss, and 1 in the second. Maximum equals to 2.
- You can get 2 in the first toss, and 2 in the second. Maximum equals to 2.
The probability of each outcome is 0.25, that is expectation equals to:
You can read about expectation using the following link: http://en.wikipedia.org/wiki/Expected_value
思路：
刚开始沙雕了。把题目中的几次投掷的最大值看成了总和，也就是投掷几次的得到的总面值，开始枚举排列，找规律，就在感觉有点眉目时，发现不对，题目好像不是这个，(⊙o⊙)…
题目的意思是投掷一枚m面的色子n次，每次投掷的面值有个最大值，求所有可能的投掷的最大期望。
开始的思路：
最大为一，一种情况，
最大为2，可以选择，有一次投了2,剩下投了少于2，两次投了2，剩下少于2，...，n次全投2
最大为3，等等等等
最后公式是：\(\sum_{i=1}^{m}\sum_{j=1}^{n}i*C_n^j(i-1)^{n-j}\)，其中i是投掷的最大面值，j是投掷中几次投到最大值。
一看，这个公式，怎么化简啊？一看不知道，再看吓一跳：
\(\sum_{j=1}^{n}C_n^j(i-1)^{n-j}=C_n^1(i-1)^{n-1}+C_n^2(i-1)^{n-2}+...+C_n^n(i-1)^0\)是不是有点像二项式的展开式啊。
是什么的展开呢？
很显然其中一项是\((i-1)\)，还有一项是1，所以想到了\((i-1)^n\)，但不对啊，\((i-1)^n=C_n^0i^n(-1)^n+C_n^1i^1(-1)^{n-1}+...+C_n^ni^n\)。诶。。。推不走了，要是直接套用公式复杂度直接\(O(n^2)\)，m和n都是\(10^5\)量级的，一秒根本不够。
只有另找公式了。
换个思路：既然是每次最大面值为i，那我投n次不超过i的数目是\(i^n\)，每次投不超过i-1的数目是\((i-1)^n\)，两者相减得到什么？就是最大为i的数目！还是必有为面值是i的情况，因为已经减了最大是i-1即以下的所有可能了。这样\(O(n)\)内可以算出结果。这里注意的是用pow函数因为参数是相邻的，可以把一个参数存起来供下一次迭代使用，从而减少计算量。还为了不让溢出，最后要总和除以\(m^n\)，变成每次除以\(m^n\)再求和。
即\(\frac{pow(i,n)-pow(i-1,n)}{m^n}=\frac{pow(i/m,n)-pow((i-1)/m,n)}{1}\)。
对了顺带还证明了：\(\sum_{i=1}^{m}\sum_{j=1}^{n}i*C_n^j(i-1)^{n-j}=\sum_{i=1}^{m}i*(i^n-(i-1)^n)\)。
代码：
#include <iomanip>
using namespace std;
double n,m;
double ans = 0;
int main()
{
cin >> m >> n;
double tmp = 0;
double last = 0;
for(int i = 1;i<=m;i++)
{
tmp = pow(i/m,n);
ans += (tmp-last)*i;
last = tmp;
}
cout.setf(ios_base::fixed,ios_base::fixed);
cout << setprecision(12) << ans << endl;
return 0;
}
