Scout YYF I
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 5020 | Accepted: 1355 |
Description
YYF is a couragous scout. Now he is on a dangerous mission which is to penetrate into the enemy‘s base. After overcoming a series difficulties, YYF is now at the start of enemy‘s famous "mine road". This is a very long road, on which there are numbers of mines. At first, YYF is at step one. For each step after that, YYF will walk one step with a probability of p, or jump two step with a probality of 1-p. Here is the task, given the place of each mine, please calculate the probality that YYF can go through the "mine road" safely.
Input
The input contains many test cases ended with EOF.
Each test case contains two lines.
The First line of each test case is N (1 ≤ N ≤ 10) and p (0.25 ≤ p ≤ 0.75) seperated by a single blank, standing for the number of mines and the probability to walk one step.
The Second line of each test case is N integer standing for the place of N mines. Each integer is in the range of [1, 100000000].
Output
For each test case, output the probabilty in a single line with the precision to 7 digits after the decimal point.
Sample Input
1 0.5 2 2 0.5 2 4
Sample Output
0.5000000 0.2500000
题意：在一条不满地雷的路上，你现在的起点在1处。在N个点处布有地雷，1<=N<=10。地雷点的坐标范围：[1,100000000].
每次前进p的概率前进一步，1-p的概率前进1-p步。问顺利通过这条路的概率。就是不要走到有地雷的地方。
设dp[i]表示到达i点的概率，则 初始值 dp[1]=1.
很容易想到转移方程： dp[i]=p*dp[i-1]+(1-p)*dp[i-2];
但是由于坐标的范围很大，直接这样求是不行的，而且当中的某些点还存在地雷。
N个有地雷的点的坐标为 x[1],x[2],x[3]```````x[N].
我们把道路分成N段：
1~x[1];
x[1]+1~x[2];
x[2]+1~x[3];
`
`
`
x[N-1]+1~x[N].
这样每一段只有一个地雷。我们只要求得通过每一段的概率。乘法原理相乘就是答案。
具体来说，对于某一段长度为nk的线段k，设a是线段k的开头，b是线段k的结尾，nk=a-b-1，
到达a的概率设为1，到达a+1概率是p，到达a+2的概率就是Pa*(1-p)+Pa+1*p，这样就可以递推了。
由于Pam=Pam-1*p+Pam-2*(1-p)，即推的公式都是一样的，可以用矩阵乘法+快速幂来做。
问题的解可以看做Pn1*(1-p)*Pn2*(1-p)*....Pnn*(1-p)，因为最后一个陷阱要跳过去才安全。
1 #include<iostream>
2 #include<cstdio>
3 #include<cstring>
4 #include<vector>
5 #include<algorithm>
6 #include<cmath>
7 #define M(a,b) memset(a,b,sizeof(a))
8 typedef long long LL;
9
10 using namespace std;
11
12 int n;
13 double p;
14 int num[20];
15
16 struct matrix
17 {
18 double mat[2][2];
19 void init()
20 {
21 mat[0][0] = p;
22 mat[0][1] = 1-p;
23 mat[1][0] = 1;
24 mat[1][1] = 0;
25 }
26 };
27
28 matrix mamul(matrix aa,matrix bb)
29 {
30 matrix c;
31 for(int i = 0;i<2;i++)
32 {
33 for(int j = 0;j<2;j++)
34 {
35 c.mat[i][j] = 0;
36 for(int k = 0;k<2;k++)
37 c.mat[i][j]+=(aa.mat[i][k]*bb.mat[k][j]);
38 }
39 }
40 return c;
41 }
42
43 matrix mul(matrix s, int k)
44 {
45 matrix ans;
46 ans.init();
47 while(k>=1)
48 {
49 if(k&1)
50 ans = mamul(ans,s);
51 k = k>>1;
52 s = mamul(s,s);
53 }
54 return ans;
55 }
56
57 int main()
58 {
59 while(scanf("%d%lf",&n,&p)==2)
60 {
61 for(int i = 1;i<=n;i++)
62 scanf("%d",&num[i]);
63 sort(num+1,num+n+1);
64 num[0] = 0;
65 if(num[1]==1) {puts("0.0000000"); continue;}
66 matrix ans;
67 ans.init();
68 double out = 1;
69 matrix tem;
70 tem.mat[0][0] = (1-p)+p*p;
71 tem.mat[0][1] = p;
72 tem.mat[1][0] = p;
73 tem.mat[1][1] = 1;
74 int flag = 0;
75 for(int i = 1;i<=n;i++)
76 {
77 if(num[i]-num[i-1]<2)
78 {puts("0.0000000"); flag = 1; break;}
79 if(num[i]-num[i-1]-2==0)
80 out*=tem.mat[1][1];
81 else
82 {
83 ans.init();
84 ans = mul(ans,num[i]-num[i-1]-3);
85 matrix c;
86 for(int i = 0;i<2;i++)
87 {
88 for(int j = 0;j<2;j++)
89 {
90 c.mat[i][j] = 0;
91 for(int k = 0;k<2;k++)
92 c.mat[i][j]+=(ans.mat[i][k]*tem.mat[k][j]);
93 }
94 }
95 //cout<<c.mat[1][1]<<‘!‘<<endl;
96 out*=c.mat[1][1];
97 }
98 out*=(1-p);//cout<<out<<endl;
99 tem.mat[0][0] = (1-p)+p*p;
100 tem.mat[0][1] = p;
101 tem.mat[1][0] = p;
102 tem.mat[1][1] = 1;
103 }
104 if(!flag)
105 printf("%.7f\n",out);
106 }
107 return 0;
108 }