Gym 100512B Betting Fast (题意+概率)
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题意:你开始有 s 元钱,然后你要在 t 场内赚到 n 元,每次赢的概率是 p,并且要越快越好。
析:当时没注意这个条件,要越快越好,然后写概率dp,怎么看也不像是对。其实是每次赌 min(s, n-s),尽快结束,就两种决策,要么赢,要么输,
就简单了。
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #include <set> #include <queue> #include <algorithm> #include <vector> #include <map> #include <cctype> #include <cmath> #include <stack> //#include <tr1/unordered_map> #define freopenr freopen("in.txt", "r", stdin) #define freopenw freopen("out.txt", "w", stdout) using namespace std; //using namespace std :: tr1; typedef long long LL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; const double inf = 0x3f3f3f3f3f3f; const LL LNF = 0x3f3f3f3f3f3f; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 1e4 + 5; const LL mod = 10000000000007; const int N = 1e6 + 5; const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1}; const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1}; const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"}; inline LL gcd(LL a, LL b){ return b == 0 ? a : gcd(b, a%b); } int n, m; const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; inline int Min(int a, int b){ return a < b ? a : b; } inline int Max(int a, int b){ return a > b ? a : b; } inline LL Min(LL a, LL b){ return a < b ? a : b; } inline LL Max(LL a, LL b){ return a > b ? a : b; } inline bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m; } int s, t; double p, q; double dfs(int time, int s){ if(s >= n) return 1.0; if(s <= 0) return 0.0; if(time <= 0) return 0.0; double ans = 0.0; ans += dfs(time - 1, s + min(s, n-s)) * p; ans += dfs(time - 1, s - min(s, n-s)) * q; return ans; } int main(){ freopen("betting.in", "r", stdin); freopen("betting.out", "w", stdout); while(scanf("%d %d %lf %d", &n, &s, &p, &t) == 4 && n){ p /= 100.0; q = 1.0 - p; double ans = dfs(t, s); printf("%.10f\n", ans); } return 0; }
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