水了一上午。。

拿6面举例子吧,因为是投掷m次取最大,最大是1概率(1/6)^m;最大是2就可以取到(1,2)那么概率就是(1/3)^m-(1/6)^m。(当前减去上一个)

 #include<bits/stdc++.h>
#define LL long long
#define LD long double
#define N 100005
using namespace std;
inline int ra()
{
int x=,f=; char ch=getchar();
while (ch<'' || ch>'') {if (ch=='-') f=-; ch=getchar();}
while (ch>='' && ch<='') {x=x*+ch-''; ch=getchar();}
return x*f;
}
double ksm(double a, int p)
{
double sum=;
for (;p;p>>=,a*=a)
if (p&)
sum*=a;
return sum;
}
int main()
{
int n=ra(),m=ra();
double ans=;
for (int i=; i<=n; i++)
ans+=(double)i*(ksm((double)i/n,m)-ksm((double)(i-)/n,m));
printf("%.12lf",ans);
return ;
}

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