m人参加n场比赛，已知某一人各比赛的名次，求它的期望总名次（各比赛名次相加的和的排名）

显然期望dp
对于其它m-1人，令Ei,s表示第i场比赛得分为s的期望


CF 601C(Kleofáš and the n-thlon-期望dp)_#define

然后前缀和乱搞

#include<bits/stdc++.h>
using namespace std;
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
#define
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#define
#define
typedef long long ll;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
int read()
{
    int x=0,f=1; char ch=getchar();
    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
    return x*f;
} 
#define
#define
double E[MAXN][MAXN*MAXM]; 
int n,m,a[MAXN];
int main()
{
    freopen("CF601C.in","r",stdin);
//  freopen(".out","w",stdout);
    n=read(),m=read();
    For(i,n) a[i]=read();
    if (m==1) {
        cout<<"1.0000000000000000\n";
        return 0;
    }
    MEM(E)
    Rep(j,m+1) E[0][j]=1.0;
    For(i,n) {
        Fork(s,i,i*m) {
            E[i][s]=E[i-1][s-1] - ((s-m-1>=0)?E[i-1][s-m-1]:0) ;
            if (s>=a[i]) E[i][s]-=E[i-1][s-a[i]]-((s-a[i]-1>=0)?E[i-1][s-a[i]-1]:0); 
            E[i][s]/=m-1;
        }
        For(s,i*m+m) { 
            E[i][s]+=E[i][s-1];
        }
    }
    int s1=0;
    For(i,n) s1+=a[i];
    cout<<setprecision(10)<<setiosflags(ios::fixed)<<(m-1)*E[n][s1-1]+1<<endl; 
    return 0;
}