Race UVA - 12034

Disky and Sooma, two of the biggest mega minds of Bangladesh went to a far country. They ate, coded

and wandered around, even in their holidays. They passed several months in this way. But everything

has an end. A holy person, Munsiji came into their life. Munsiji took them to derby (horse racing).

Munsiji enjoyed the race, but as usual Disky and Sooma did their as usual task instead of passing some

romantic moments. They were thinking- in how many ways a race can finish! Who knows, maybe this

is their romance!

In a race there are n horses. You have to output the number of ways the race can finish. Note that,

more than one horse may get the same position. For example, 2 horses can finish in 3 ways.

1. Both first

2. horse1 first and horse2 second

3. horse2 first and horse1 second

Input

Input starts with an integer T (≤ 1000), denoting the number of test cases. Each case starts with a

line containing an integer n (1 ≤ n ≤ 1000).

Output

For each case, print the case number and the number of ways the race can finish. The result can be

very large, print the result modulo 10056.

Sample Input

3

1

2

3

Sample Output

Case 1: 1

Case 2: 3

Case 3: 13

可以有多匹马占用同一个名次，问n匹马有几种比赛结果。

状态转移方程是s[i][j]=(s[i-1][j]*j+s[i-1][j-1]*j)，指i匹马最后有j个名次的情况，可以由i-1匹有j个名次的情况中，某个名次多加一匹马得到，也可以由i-1匹有j-1个名次的情况中，多加一匹马单独占有一个名次得到，即两种情况之和。

#include <iostream>
#include <string.h>
using namespace std;

int s[1005][1005];
int ss[1005];
int main()
{
int i,j,T,n,k;

memset(s,0,sizeof(s));
memset(ss,0,sizeof(ss));
s[1][1]=1;
ss[1]=1;
for(i=2;i<=1000;i++){
for(j=1;j<=i;j++){
s[i][j]=(s[i-1][j]*j+s[i-1][j-1]*j)%10056;
ss[i]+=s[i][j];
ss[i]%=10056;
}
}
cin>>T;
k=1;
while(k<=T){
cin>>n;
cout<<"Case "<<k<<": ";
cout<<ss
<<endl;
k++;
}
return 0;
}