点击打开链接 A. 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 mdots. Twilight Sparkle is sure that when the dice is tossed, each face appears with probability 
. Also she knows that each tZ??http://www.2cto.com/kf/ware/vc/" target="_blank" class="keylink">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"t exceed 10??-?4.
Sample test(s) input
6 1
output
3.500000000000
input
6 3
output
4.958333333333
input
2 2
output
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
m面的骰子，投掷n次，得最大面为k的概率是：k^m/n^m-(k-1)^m/n^m=(k/n)^m-最大值为k-1的概率。
//31 ms 0 KB
#include
#include
int main() { double n,m; while(scanf("%lf%lf",&m,&n)!=EOF) { double a=0,c=0; for(int i=1;i<=m;i++) { double b=pow((double)(i)/m,n); a+=(b-c)*i; c=b; } printf("%.12lf\n",a); } return 0; }