UVA 12034（递推&递归_I题）解题报告

题目链接：https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=3185

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题意：赛马结果，求有多少种

思路：递推关系：dp[i][j]=j*(dp[i-1][j](第i个马和前面的马搭伙到达）+dp[i-1][j-1]（第i个马单独算一次））

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<string>
#include<vector>
#include<stack>
#include<bitset>
#include<cstdlib>
#include<cmath>
#include<set>
#include<list>
#include<deque>
#include<map>
#include<queue>
using namespace std;
typedef long long ll;
const double PI = acos(-1.0);
const double eps = 1e-6;
const int MAXN =1e3+10;
int ans[MAXN];
int h[MAXN][MAXN]={0};
int T;
const int MOD =10056;

int horse(){
    for(int i=1;i<MAXN;i++){
        h[i][1]=1;
        for(int j=2;j<=i;j++){
            h[i][j]=(h[i-1][j-1]*j%MOD+h[i-1][j]*j%MOD)%MOD;
        }
        for(int j=1;j<=i;j++){
            ans[i]=(ans[i]+h[i][j])%MOD;
        }

    }
    return 0;

}

int main(void){
    horse();
    scanf("%d",&T);
    //freopen("out.txt","wa",stdout);
    for(int i=0;i<T;i++){
        int a=0;
        scanf("%d",&a);

        printf("Case %d: %d\n",i+1,ans[a]);
    }
    return 0;

}

原文地址：https://www.cnblogs.com/caomingpei/p/8352320.html