Codeforces Round #259 (Div. 2) C Little Pony and Expected Maximum

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用逆向思维解+容斥原理

最大为 m 个的概率为 1减去不是m最大的概率，(1 - (( m-1 )/ m ) ^ n )，则 m-1 的概率为在不是 m 的状态下乘以 1减去不是m-1最大的概率

以此类推

感谢楠哥的思路

#include <algorithm>#include <iostream>#include <iomanip>#include <cstring>#include <climits>#include <complex>#include <fstream>#include <cassert>#include <cstdio>#include <bitset>#include <vector>#include <deque>#include <queue>#include <stack>#include <ctime>#include <set>#include <map>#include <cmath>#define eps 1e-9#define INF 0x3f3f3f3fusing namespace std;typedef long long ll;typedef long double ld;typedef pair<ll, ll> pll;typedef complex<ld> point;typedef pair<int, int> pii;typedef pair<pii, int> piii;template<class T>inline bool read(T &n){    T x = 0, tmp = 1; char c = getchar();    while((c < '0' || c > '9') && c != '-' && c != EOF) c = getchar();    if(c == EOF) return false;    if(c == '-') c = getchar(), tmp = -1;    while(c >= '0' && c <= '9') x *= 10, x += (c - '0'),c = getchar();    n = x*tmp;    return true;}template <class T>inline void write(T n){    if(n < 0)    {        putchar('-');        n = -n;    }    int len = 0,data[20];    while(n)    {        data[len++] = n%10;        n /= 10;    }    if(!len) data[len++] = 0;    while(len--) putchar(data[len]+48);}//-----------------------------------const int MAXN=100010;int n,m;double ans=0;int main(){    int n,m;    read(m);read(n);    for (int i=1;i<=m;i++)        ans+=i*pow((double)i/(double)m,n)*(1-pow((double)(i-1)/(double)i,n));    printf("%.12lf",ans);    return 0;}

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