UVA - 10081 Tight Words

最新推荐文章于 2018-03-14 16:37:07 发布

weixin_30918633

最新推荐文章于 2018-03-14 16:37:07 发布

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原文链接：http://www.cnblogs.com/mofushaohua/p/7789465.html

/*
  这题的dp思路挺巧妙，dp[i][j]记录当长度为i时，末位为j的，并且满足题目要求的，取法...并且还除以了截至该位，占所有排列的概率(每次更新dp，都有除以(1+k) )
  
  解析见：
  http://blog.youkuaiyun.com/codebattle/article/details/38420827
  http://www.cnblogs.com/scau20110726/archive/2013/02/17/2914763.html
*/

#include <iostream>
#include <iomanip>
using namespace std;
int k, n;
const int MAXN = 110;
const int MAXM = 15;
double dp[MAXN][MAXM];
void solve()
{
	for (int i = 0; i <= k; i++)
	dp[1][i] = 100.0 / (k + 1);
	
	for (int i = 2; i <= n; i++)
	for (int j = 0; j <= k; j++)
	{
		dp[i][j] = dp[i - 1][j] / (k + 1);
		if (j != 0)
		dp[i][j] += dp[i - 1][j - 1] / (k + 1);
		
		if (j != k)
		dp[i][j] += dp[i - 1][j + 1] / (k + 1);
		
	}
	
	double ans = 0;
	for (int i = 0; i <= k; i++)
	ans += dp[n][i];
	
	cout << fixed << setprecision(5) << ans << endl;
	
	
}
int main()
{
	while (cin >> k >> n)
	{
		solve();
	}
	return 0;
}

转载于:https://www.cnblogs.com/mofushaohua/p/7789465.html