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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;}

 

Gym 100512B Betting Fast (题意+概率)