DIV2 1000pt
题意：一群人排队，每次操作由要骰子决定，只要没有人中奖，游戏就不结束。若摇骰子摇出4，则队列第一个人中奖；否则，若摇的是奇数，则第一个人排队到队伍末尾去；否则，第一个人出局。若游戏途中，队列只剩一个人，则直接中奖。若摇了k次骰子仍然没人中奖，则此时队列第一个人自动获奖，游戏结束。给出k，问队列初始时为n个人，此时排在第m个的人中奖的概率有多大。n, m, k <= 10。
解法：普通的概率dp。将题中的k记为num。
法一：设d[i][j][k]表示当前状态为摇了i次骰子，队列共有j人，排在第k个的中奖的概率。具体状态转移方程见代码，很简单。
法二：设d[i][j][k]表示由初始状态变成摇了i次骰子后，队列有j个人，初始时排在第m个的人现在排在第k个的概率。状态转移方程同见方程。
tag:概率dp
法一：
1 // BEGIN CUT HERE 2 /* 3 * Author: plum rain 4 * score : 5 */ 6 /* 7 8 */ 9 // END CUT HERE 10 #line 11 "TheTicketsDivTwo.cpp" 11 #include <sstream> 12 #include <stdexcept> 13 #include <functional> 14 #include <iomanip> 15 #include <numeric> 16 #include <fstream> 17 #include <cctype> 18 #include <iostream> 19 #include <cstdio> 20 #include <vector> 21 #include <cstring> 22 #include <cmath> 23 #include <algorithm> 24 #include <cstdlib> 25 #include <set> 26 #include <queue> 27 #include <bitset> 28 #include <list> 29 #include <string> 30 #include <utility> 31 #include <map> 32 #include <ctime> 33 #include <stack> 34 35 using namespace std; 36 37 #define CLR(x) memset(x, 0, sizeof(x)) 38 #define CLR1(x) memset(x, -1, sizeof(x)) 39 #define PB push_back 40 #define SZ(v) ((int)(v).size()) 41 #define zero(x) (((x)>0?(x):-(x))<eps) 42 #define out(x) cout<<#x<<":"<<(x)<<endl 43 #define tst(a) cout<<#a<<endl 44 #define CINBEQUICKER std::ios::sync_with_stdio(false) 45 46 typedef vector<int> VI; 47 typedef vector<string> VS; 48 typedef vector<double> VD; 49 typedef long long int64; 50 51 const double eps = 1e-8; 52 const double PI = atan(1.0)*4; 53 const int maxint = 2139062143; 54 55 double d[20][20][20]; 56 57 class TheTicketsDivTwo 58 { 59 public: 60 double find(int n, int m, int num){ 61 double t1 = (double) 1 / (double)3, t2 = (double)1 / (double)6; 62 CLR (d); 63 for (int i = 1; i <= n; ++ i) 64 d[num][i][1] = 1; 65 66 for (int i = num-1; i >= 0; -- i) 67 for (int j = 1; j <= n; ++ j) 68 for (int k = 1; k <= j; ++ k){ 69 if (j == 1) d[i][j][k] = 1; 70 else if (k == 1) d[i][j][k] = d[i+1][j][j]*0.5 + t2; 71 else d[i][j][k] = d[i+1][j][k-1]*0.5 + d[i+1][j-1][k-1]*t1; 72 } 73 return d[0][n][m]; 74 } 75 76 // BEGIN CUT HERE 77 public: 78 void run_test(int Case) { if ((Case == -1) || (Case == 0)) test_case_0(); if ((Case == -1) || (Case == 1)) test_case_1(); if ((Case == -1) || (Case == 2)) test_case_2(); if ((Case == -1) || (Case == 3)) test_case_3(); } 79 private: 80 template <typename T> string print_array(const vector<T> &V) { ostringstream os; os << "{ "; for (typename vector<T>::const_iterator iter = V.begin(); iter != V.end(); ++iter) os << '"' << *iter << "","; os << " }"; return os.str(); } 81 void verify_case(int Case, const double &Expected, const double &Received) { cerr << "Test Case #" << Case << "..."; if (Expected == Received) cerr << "PASSED" << endl; else { cerr << "FAILED" << endl; cerr << " Expected: "" << Expected << '"' << endl; cerr << " Received: "" << Received << '"' << endl; } } 82 void test_case_0() { int Arg0 = 2; int Arg1 = 1; int Arg2 = 1; double Arg3 = 0.16666666666666666; verify_case(0, Arg3, find(Arg0, Arg1, Arg2)); } 83 void test_case_1() { int Arg0 = 2; int Arg1 = 1; int Arg2 = 2; double Arg3 = 0.5833333333333334; verify_case(1, Arg3, find(Arg0, Arg1, Arg2)); } 84 void test_case_2() { int Arg0 = 7; int Arg1 = 7; int Arg2 = 4; double Arg3 = 0.0; verify_case(2, Arg3, find(Arg0, Arg1, Arg2)); } 85 void test_case_3() { int Arg0 = 4; int Arg1 = 2; int Arg2 = 10; double Arg3 = 0.25264033564814814; verify_case(3, Arg3, find(Arg0, Arg1, Arg2)); } 86 87 // END CUT HERE 88 89 }; 90 91 // BEGIN CUT HERE 92 int main() 93 { 94 // freopen( "a.out" , "w" , stdout ); 95 TheTicketsDivTwo ___test; 96 ___test.run_test(-1); 97 return 0; 98 } 99 // END CUT HERE
法二：
1 // BEGIN CUT HERE 2 /* 3 * Author: plum rain 4 * score : 5 */ 6 /* 7 8 */ 9 // END CUT HERE 10 #line 11 "TheTicketsDivTwo.cpp" 11 #include <sstream> 12 #include <stdexcept> 13 #include <functional> 14 #include <iomanip> 15 #include <numeric> 16 #include <fstream> 17 #include <cctype> 18 #include <iostream> 19 #include <cstdio> 20 #include <vector> 21 #include <cstring> 22 #include <cmath> 23 #include <algorithm> 24 #include <cstdlib> 25 #include <set> 26 #include <queue> 27 #include <bitset> 28 #include <list> 29 #include <string> 30 #include <utility> 31 #include <map> 32 #include <ctime> 33 #include <stack> 34 35 using namespace std; 36 37 #define CLR(x) memset(x, 0, sizeof(x)) 38 #define CLR1(x) memset(x, -1, sizeof(x)) 39 #define PB push_back 40 #define SZ(v) ((int)(v).size()) 41 #define zero(x) (((x)>0?(x):-(x))<eps) 42 #define out(x) cout<<#x<<":"<<(x)<<endl 43 #define tst(a) cout<<#a<<endl 44 #define CINBEQUICKER std::ios::sync_with_stdio(false) 45 46 typedef vector<int> VI; 47 typedef vector<string> VS; 48 typedef vector<double> VD; 49 typedef long long int64; 50 51 const double eps = 1e-8; 52 const double PI = atan(1.0)*4; 53 const int maxint = 2139062143; 54 55 double d[20][20][20]; 56 57 class TheTicketsDivTwo 58 { 59 public: 60 double find(int n, int m, int num){ 61 double t1 = (double) 1 / (double)3, t2 = (double)1 / (double)6; 62 CLR (d); 63 d[0][n][m] = 1.0; 64 for (int i = 1; i <= num; ++ i) 65 for (int j = 1; j <= n; ++ j) 66 for (int k = 1; k <=j; ++ k){ 67 if (j == k){ 68 if (j == 1) d[i][1][1] = d[i-1][j+1][j+1] * t1; 69 else d[i][j][k] = d[i-1][j][1] * 0.5 + d[i-1][j+1][j+1] * t1; 70 } 71 else{ 72 d[i][j][k] = d[i-1][j][k+1] * 0.5; 73 if (j < n) d[i][j][k] += d[i-1][j+1][k+1] * t1; 74 } 75 } 76 77 double ans = 0; 78 for (int i = 0; i < num; ++ i) 79 for (int j = 1; j <= n; ++ j) 80 ans += d[i][j][1] * (j == 1 ? 1 : t2); 81 for (int i = 1; i <= n; ++ i) 82 ans += d[num][i][1]; 83 return ans; 84 } 85 86 // BEGIN CUT HERE 87 public: 88 void run_test(int Case) { if ((Case == -1) || (Case == 0)) test_case_0(); if ((Case == -1) || (Case == 1)) test_case_1(); if ((Case == -1) || (Case == 2)) test_case_2(); if ((Case == -1) || (Case == 3)) test_case_3(); } 89 private: 90 template <typename T> string print_array(const vector<T> &V) { ostringstream os; os << "{ "; for (typename vector<T>::const_iterator iter = V.begin(); iter != V.end(); ++iter) os << '"' << *iter << "","; os << " }"; return os.str(); } 91 void verify_case(int Case, const double &Expected, const double &Received) { cerr << "Test Case #" << Case << "..."; if (Expected == Received) cerr << "PASSED" << endl; else { cerr << "FAILED" << endl; cerr << " Expected: "" << Expected << '"' << endl; cerr << " Received: "" << Received << '"' << endl; } } 92 void test_case_0() { int Arg0 = 2; int Arg1 = 1; int Arg2 = 1; double Arg3 = 0.16666666666666666; verify_case(0, Arg3, find(Arg0, Arg1, Arg2)); } 93 void test_case_1() { int Arg0 = 2; int Arg1 = 1; int Arg2 = 2; double Arg3 = 0.5833333333333334; verify_case(1, Arg3, find(Arg0, Arg1, Arg2)); } 94 void test_case_2() { int Arg0 = 7; int Arg1 = 7; int Arg2 = 4; double Arg3 = 0.0; verify_case(2, Arg3, find(Arg0, Arg1, Arg2)); } 95 void test_case_3() { int Arg0 = 4; int Arg1 = 2; int Arg2 = 10; double Arg3 = 0.25264033564814814; verify_case(3, Arg3, find(Arg0, Arg1, Arg2)); } 96 97 // END CUT HERE 98 99 }; 100 101 // BEGIN CUT HERE 102 int main() 103 { 104 // freopen( "a.out" , "w" , stdout ); 105 TheTicketsDivTwo ___test; 106 ___test.run_test(-1); 107 return 0; 108 } 109 // END CUT HERE