poj 3744 Scout (Another) YYF I - 概率与期望 - 动态规划 - 矩阵快速幂

  1 /**
  2  * poj
  3  * Problem#3744
  4  * Accepted
  5  * Time: 0ms
  6  * Memory: 232k 
  7  */
  8 #include <iostream>
  9 #include <fstream>
 10 #include <sstream>
 11 #include <cstdio>
 12 #include <cstdlib>
 13 #include <cstring>
 14 #include <cmath>
 15 #include <ctime>
 16 #include <cctype>
 17 #include <algorithm>
 18 #include <set>
 19 #include <map>
 20 #include <vector>
 21 #include <queue>
 22 #include <stack>
 23 #include <bitset>
 24 #include <cassert>
 25 #ifndef WIN32
 26 #define Auto "%lld"
 27 #else
 28 #define Auto "%I64d"
 29 #endif
 30 using namespace std;
 31 typedef bool boolean;
 32 #define clra(a) memset(a, false, sizeof(a))
 33 const signed int inf = ((~0u) >> 1);
 34 #define smin(a, b) a = min(a, b)
 35 #define smax(a, b) a = max(a, b)
 36 
 37 template <typename T>
 38 class Matrix {
 39     public:
 40         T a[2][2];
 41         int row; 
 42         int col;
 43         
 44         Matrix() {        }
 45         Matrix(int row, int col):row(row), col(col) {        }
 46         
 47         Matrix<T> operator * (Matrix<T> b) {
 48             Matrix<T> rt(row, b.col);
 49             assert(col == b.row);
 50             for(int i = 0; i < row; i++)
 51                 for(int j = 0; j < b.col; j++) {
 52                     rt[i][j] = 0;
 53                     for(int k = 0; k < col; k++)
 54                         rt[i][j] += a[i][k] * b[k][j];
 55                 }
 56             return rt;
 57         } 
 58         
 59         T* operator [] (int pos) {
 60             return &a[pos][0];
 61         }
 62 };
 63 
 64 int n;
 65 double p;
 66 Matrix<double> unit(2, 2);
 67 Matrix<double> trans(2, 2);
 68 int boom[15];
 69 
 70 template <typename T>
 71 Matrix<T> matpow(Matrix<T>& a, int pos) {
 72     if(pos == 0)    return unit;
 73     if(pos == 1)    return a;
 74     Matrix<T> temp = matpow(a, pos >> 1);
 75     temp = temp * temp;
 76     if(pos & 1)    temp = temp * a;
 77     return temp;
 78 }
 79 
 80 inline void prepare() {
 81     unit[0][0] = unit[1][1] = 1;
 82     unit[0][1] = unit[1][0] = 0;
 83 }
 84 
 85 inline void init() {
 86     for(int i = 0; i < n; i++)
 87         scanf("%d", boom + i);
 88     sort(boom, boom + n);
 89     trans[0][0] = p, trans[0][1] = 1 - p;
 90     trans[1][0] = 1, trans[1][1] = 0;
 91 }
 92 
 93 inline void solve() {
 94     int last = boom[n - 1];
 95     Matrix<double> f(2, 1);
 96     f[0][0] = 0, f[1][0] = 1;
 97     for(int i = n - 2; ~i; f[0][0] = 0, last = boom[i], i--)
 98         f = matpow(trans, last - boom[i]) * f;
 99     f = matpow(trans, last - 1) * f;
100     printf("%.7lf\n", f[0][0]);
101 }
102 
103 int main() {
104     prepare();
105     while(~scanf("%d%lf", &n, &p)) {    //坑啊，scanf交G++ WA了，交C++过了 
106         init();
107         solve();
108     }
109     return 0;
110 }