Scout YYF I
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 9552 | Accepted: 2793 |
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].
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
0.5 0.5
Sample Output
0.5000000
0.2500000
分析：
概率dp是需要顺着推的，定义dp[i]表示到i的概率。当i有雷时，dp[i] = 0；
否则 dp[i] = dp[i - 1] * p + dp[i - 1] * （1 - p）
最后输出dp[dis[n] + 1]即行。
因为n很大，转移又固定，可以联想到矩乘 + 快速幂。于是就0MS过了
坑： 输入雷不一定有序，需要排序，精度输出需要%.7f 不能 %.7lf。
AC代码：
# include <iostream>
# include <cstdio>
# include <cstring>
# include <algorithm>
using namespace std;
int n;
int dis[];
struct fi{
double data[][];
fi(){
for(int i = ;i < ;i++){
for(int j = ;j < ;j++){
data[i][j] = ;
}
}
}
}A,T,O;
inline fi operator *(fi A,fi B){
fi t;
for(int i = ;i < ;i++){
for(int j = ;j < ;j++){
for(int k = ;k < ;k++){
t.data[i][j] += A.data[i][k] * B.data[k][j];
}
}
}
return t;
}
fi cmd(fi C,int k){
fi D = O;
while(k){
if(k & )D = D * C;
k >>= ;
C = C * C;
}
return D;
}
double p;
int main(){
for(int i = ;i < ;i++)O.data[i][i] = 1.0;
while(~scanf("%d %lf",&n,&p)){
T.data[][] = (1.0 - p);
T.data[][] = p;
T.data[][] = 1.0;
T.data[][] = 0.0;
for(int i = ;i < ;i++){
for(int j = ;j < ;j++){
A.data[i][j] = 0.0;
}
}
A.data[][] = 1.0;
for(int i = ;i <= n;i++){
scanf("%d",&dis[i]);
}
sort(dis + ,dis + n + );
for(int i = ;i <= n;i++){
A = A * cmd(T,dis[i] - dis[i - ] - );
A.data[][] = ;
A = A * T;
}
printf("%.7f\n",A.data[][]);
}
}