UVA 10081 Tight numbers(POJ 2537)

　　直接看代码就OK。思路比较简单。就是注意概率要在转移过程中算出来。不能算成成立的方案书除以总方案数（POJ的这道题可以这么干。数据很水么。另外POJ要用%.5f,%.5lf 会WA。）

#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <stack>
#include <queue>
#include <cctype>
#include <cstdio>
#include <string>
#include <vector>
#include <climits>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#define LL long long
#define PI 3.1415926535897932626
using namespace std;
int gcd(int a, int b) {return a % b ==  ? b : gcd(b, a % b);}
double dp[][];
int N,K;
void slove()
{
    if (K <= ) {puts("100.00000");return;}
    for (int i =  ;i < ; i++) for (int j = ;j <; j++) dp[i][j]=0.0;
    for (int i =  ;i <= K; i++) dp[][i]=100.0/(double)(K+);
    for (int i = ; i <= N; i++)
    {
        dp[i][] = 1.0/(double)(K+) * ( dp[i-][] + dp[i-][]);
        for (int j = ; j <= K ; j++)
        {
            if (j == K) dp[i][j] = 1.0/(double)(K+) * (dp[i-][K] + dp[i-][K-]);
            else  dp[i][j] = 1.0/(double)(K+) * (dp[i-][j-] + dp[i-][j] + dp[i-][j+]);
        }
    }
    double ans=0.0;
    for (int i =  ;i <= K; i ++) ans+=dp[N][i];
    printf("%.5lf\n",ans);
}
int main()
{
    while (scanf("%d%d",&K,&N)!=EOF)
    slove();
    return ;
}