例题10-14 比赛名次 UVa12034

1.题目描述：点击打开链接

2.解题思路：本题通过寻找递推关系解决。设最终答案是f(n)。假设第一名有i个人，有C(n,i)种可能，接下来有f(n-i)种可能性。最终答案是∑C(n,i)f(n-i)(i从1开始)。

3.代码：

#define _CRT_SECURE_NO_WARNINGS
#include<iostream>
#include<algorithm>
#include<string>
#include<sstream>
#include<set>
#include<vector>
#include<stack>
#include<map>
#include<queue>
#include<deque>
#include<cstdlib>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<ctime>
#include<functional>
using namespace std;

#define N 1000+10
#define MOD 10056
int C[N][N];
int f[N];
int n;
void init()
{
	C[0][0] = f[0] = 1;
	for (int i = 1; i <= 1000; i++)
	{
		C[i][0] = C[i][i]=1;
		f[i] = 0;
		for (int j = 1; j < i; j++)
			C[i][j] = (C[i - 1][j - 1] + C[i - 1][j])%MOD;
		for (int j = 1; j <= i; j++)
			f[i] = (f[i] + C[i][j] * f[i - j]%MOD) % MOD;
	}
}
int main()
{
	freopen("test.txt", "r", stdin);
	int cnt = 0;
	init();
	int t;
	cin >> t;
	while (t--)
	{
		cin >> n;
		printf("Case %d: ", ++cnt);
		cout << f[n] << endl;
	}
	return 0;
}