递推公式，假设第一名有i个人并列，那么:

f[n] = C(n,i) * f[n - i];

打出1 ~ 1000的所有组合数，之后记忆化搜索，需要打表。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<vector>
#include<stack>
#include<queue>
#include<map>
#include<set>
#include<list>
#include<cmath>
#include<string>
#include<sstream>
#include<ctime>
using namespace std;
#define _PI acos(-1.0)
#define INF (1 << 10)
#define esp 1e-6
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> pill;
/*===========================================
===========================================*/
#define MAXD 1000 + 10
#define mod  10056
LL f[MAXD];
LL C[MAXD][MAXD];
LL _C(){
    C[1][1] = 1; C[1][0] = 1;
    for(int i = 2 ; i <= MAXD - 10 ; i++)
        for(int j = 0 ; j <= i ; j++){
            if(j == 0)
                C[i][j] = 1;
            else
                C[i][j] = C[i - 1][j - 1] + C[i - 1][j];
            C[i][j] %= mod;
        }
}
LL _f(int n){
    LL ans = 0;
    if(f[n] != -1)
        return f[n];
    for(int i = 1 ; i <= n ; i++){
        ans = ans + C[n][i] * _f(n - i);
        ans %= mod;
    }
    f[n] = ans;
    return f[n];
}
int main(){
    memset(f,-1,sizeof(f));
    f[1] = 1; f[0] = 1;
    _C();
    _f(MAXD - 10);
    int T,Case = 1;
    scanf("%d",&T);
    while(T--){
        int n;
        scanf("%d",&n);
        printf("Case %d: %lld\n",Case++,f[n]);
    }
    return 0;
}