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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
    ", ans);
        }
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/dwtfukgv/p/5924975.html
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