Aeroplane chess

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1581    Accepted Submission(s): 1082

Problem Description

Hzz loves aeroplane chess very much. The chess map contains N+1 grids labeled from 0 to N. Hzz starts at grid 0. For each step he throws a dice(a dice have six faces with equal probability to face up and the numbers on the faces are 1,2,3,4,5,6). When Hzz is at grid i and the dice number is x, he will moves to grid i+x. Hzz finishes the game when i+x is equal to or greater than N. There are also M flight lines on the chess map. The i-th flight line can help Hzz fly from grid Xi to Yi (0<Xi<Yi<=N) without throwing the dice. If there is another flight line from Yi, Hzz can take the flight line continuously. It is granted that there is no two or more flight lines start from the same grid. Please help Hzz calculate the expected dice throwing times to finish the game.

Input

There are multiple test cases. Each test case contains several lines. The first line contains two integers N(1≤N≤100000) and M(0≤M≤1000). Then M lines follow, each line contains two integers Xi,Yi(1≤Xi<Yi≤N).   The input end with N=0, M=0.

Output

For each test case in the input, you should output a line indicating the expected dice throwing times. Output should be rounded to 4 digits after decimal point.

Sample Input

2 0 8 3 2 4 4 5 7 8 0 0

Sample Output

1.1667 2.3441

Source

2012 ACM/ICPC Asia Regional Jinhua Online

有一个长条形棋盘，每一次可以跳1,2,3,4,5,6中的一种的距离，且每一跳的概率相等1/6；且在某一个点出有一个航班可以帮助他不需要跳直接可以达到y处

问到n平均需要跳多少次。

    dp【n】=0;因为在哪里不需要再跳，这个已知，所以可以倒退

    dp[i]=sum{1+dp[n+j]}(1<=j<=6)

 代码：

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<stack>
 4 #define maxn 100008
 5 using namespace std;
 6 double  dp[maxn];
 7 stack<int> path[maxn];
 8 int pa[maxn];
 9 int main()
10 {
11   int n,m,a,b;
12   while(scanf("%d%d",&n,&m)!=EOF&&n+m)
13   {
14     memset(pa,0,sizeof(int)*(n+2));
15     memset(dp,0,sizeof(double)*(n+8));
16       while(m--)
17     {
18       scanf("%d%d",&a,&b);
19       path[b].push(a);
20       pa[a]=b;
21     }
22     while(--n>=0)
23     {
24       while(!path[n+1].empty())
25      {
26          dp[path[n+1].top()]=dp[n+1];
27          path[n+1].pop();
28      }
29       if(pa[n]==0)
30          dp[n]=1.0+(dp[n+1]+dp[n+2]+dp[n+3]+dp[n+4]+dp[n+5]+dp[n+6])/6.0;
31     }
32    printf("%.4lfn",dp[0]);
33   }
34   return 0;
35 }