Race

本文链接：https://blog.youkuaiyun.com/u014733623/article/details/38008185

本文探讨了如何解决一个关于赛马比赛排列组合的问题，通过递推算法计算不同数量马匹参赛时比赛可能的结果总数。利用数学原理和编程技巧，详细解释了解决该问题的方法和步骤。

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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.

Both first
horse₁ first and horse₂ second
horse₂ first and horse₁ second

Input

Input starts with an integer T ( $\le$ 1000 ), denoting the number of test cases. Each case starts with a line containing an integer n ( 1 $\le$ n $\le$ 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

Sample Output

Case 1: 1
Case 2: 3
Case 3: 13

题意：有n匹马，每次可以乘至少一匹，问有多少种情况。（马与马不同）。

思路：递推，num[i][j]表示i匹马，一共乘j次，num[i][j]=(num[i-1][j]*j+j*num[i-1][j-1])%mod;前一个的意思是i-1匹马乘j次后，第i匹马可以放在任意次去乘，后一个的意思是第i匹马被单独乘，可以有j种单独的方式。

AC代码如下：

#include<cstdio>
#include<cstring>
using namespace std;
int num[1100][1100],mod=10056,ans[1100];
int main()
{ int i,j,k,n,t;
  num[1][1]=1;
  for(i=2;i<=1000;i++)
   for(j=1;j<=i;j++)
    num[i][j]=(num[i-1][j]*j+j*num[i-1][j-1])%mod;
  for(i=1;i<=1000;i++)
   for(j=1;j<=i;j++)
    ans[i]=(ans[i]+num[i][j])%mod;
  scanf("%d",&t);
  for(i=1;i<=t;i++)
  { scanf("%d",&n);
    printf("Case %d: %d\n",i,ans[n]);
  }
}