codeforces 453A

最新推荐文章于 2018-11-16 11:57:44 发布

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本文介绍了一个关于计算掷特殊骰子多次后的期望最大值的问题。骰子有m面，每面点数从1递增到m，每次掷骰子各面出现的概率相同且独立。文章提供了一段C++代码，用于计算给定点数m和掷骰次数n时的期望最大值。

A. 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
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:
水概率不多bb

#include<iostream>
#include<queue>
#include<cmath>
#include<cstdio>
#include<algorithm>
using namespace std;
double ksm(double ds, int zs)
{
    double fs = 1;
    while (zs)
    {
        if (zs & 1)fs *= ds;
        ds *= ds;
        zs >>= 1;
    }
    return fs;
}
double tu[100001];
int main()
{
    int m, n;
    cin >> m >> n;
    for (int a = 1; a <= m; a++)
    {
        tu[a] = ksm(a / (double)m, n);
    }
    double ss = 0;
    for (int a = 1; a <= m; a++)
    {
        ss += a*(tu[a] - tu[a - 1]);
    }
    printf("%.9f", ss);
}