uva 12034 Race（dp）

  Race 

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 ( 1n1000).

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 

3123

Sample Output 

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

容易找到dp[i][j]表示i个数种选j个相同的方案数，转移方程详见代码。

#include <cstdio>#include <algorithm>#include <vector>#include <map>#include <queue>#include <iostream>#include <stack>#include <set>#include <cstring>#include <stdlib.h>#include <cmath>using namespace std;typedef long long LL;typedef pair<int, int> P;const int maxn = 1000 + 5;const int mod = 10056;int dp[maxn][maxn];void solve(){    memset(dp, 0, sizeof(dp));    dp[1][0] = dp[1][1] = 1;    for(int i = 2;i < maxn;i++){        dp[i][0] = (dp[i-1][0]*i) % mod;        dp[i][i] += dp[i][0];        for(int j = 1;j < i-1;j++){            dp[i][j] = ((i-j)*(dp[i-1][j] + dp[i-1][j-1])) % mod;            dp[i][i] += dp[i][j];        }        dp[i][i-1] = 1;        dp[i][i] = (dp[i][i] + dp[i][i-1]) % mod;    }}int main(){    int t,kase = 0;    solve();    scanf("%d", &t);    while(t--){        kase++;        int n;        scanf("%d", &n);        printf("Case %d: %d\n", kase, dp[n][n]);    }    return 0;}