UVA

/*  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.csdn.net/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;}