Codeforces Round #259 (Div. 1) A. Little Pony and Expected Maximum
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一个数学期望的题目,有dp的思想。但其实数学好的可以直接推公式。代码如下:
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
int main()
{
double i , n , m ;
while(~scanf("%lf %lf", &m, &n))
{
double sum = 0.0 ;
for(i=1;i<=m;i++)
{
sum+=i*(pow((i/m),n)-pow((i-1)/m,n));
}
printf("%lf\n", sum);
}
return 0;
}
这里说明一下公式的由来:
假设有6个面,投掷6次
4*(pow(4/6,6)-pow(3/6,6)):表示(面数不超过4的情况--不超过3的情况)*面数
总结:好好学习数学
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