UVA 12034 Race （遞推神馬的），12034神馬

UVA 12034 Race （遞推神馬的），12034神馬

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 n1000).

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

對，又是遞推題目：我們dp[i][j]表示i個馬，分j次到達的方法；

考慮狀態轉移：dp[i][j]=j*(dp[i-1][j](第i個馬和前面的馬搭夥到達）+dp[i-1][j-1]（第i個馬單獨算一次）)

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<limits.h>typedef long long LL;using namespace std;const int MOD=10056;int dp[1100][1100];int ans[1100];void init(){    memset(dp,0,sizeof(dp));    memset(ans,0,sizeof(ans));    dp[0][0]=1;    for(int i=1;i<=1000;i++)    {        int sum=0;        for(int j=1;j<=i;j++)        {            dp[i][j]+=(dp[i-1][j]+dp[i-1][j-1])%MOD*j%MOD;//小心心爆int            sum=(sum%MOD+dp[i][j])%MOD;        }        ans[i]=sum;    }}int main(){    init();    int n,t;    int cas=1;    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        printf("Case %d: %d\n",cas++,ans[n]);    }    return 0;}