Gym 100512B Betting Fast (题意+概率)

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;
}