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题目链接:https://vjudge.net/problem/ZOJ-3329


#include<cstdio>
#include<cstring>
#include<string>
#include<cctype>
#include<iostream>
#include<set>
#include<map>
#include<cmath>
#include<sstream>
#include<vector>
#include<stack>
#include<queue>
#include<algorithm>
#define fin freopen("a.txt","r",stdin)
#define fout freopen("a.txt","w",stdout)
typedef long long LL;
using namespace std;
typedef pair<int, int> P;
const int INF = 1e8 + 10;
const int maxn = 1000 + 10;
double A[maxn], B[maxn], p[maxn], p0;
int n, k1, k2, k3, a, b, c;

int main()
{
	//fin;
	int T;
	scanf("%d", &T);
	while(T--)
	{
		cin >> n >> k1 >> k2 >> k3 >> a >> b >> c;
        p0 = 1.0/(k1*k2*k3);
        memset(p, 0, sizeof p);
        memset(A, 0, sizeof A);
        memset(B, 0, sizeof B);
        for(int i = 1; i <= k1; i++)
          for(int j = 1; j <= k2; j++)
          	for(int k = 1; k <= k3; k++)
          		if(i != a || j != b || k != c)
          		   p[i+j+k] += p0;
        for(int i = n; i >= 0; i--)
        {
        	A[i] = p0; B[i] = 1.0;
        	for(int k = 3; k + i <= n; k++)
        	{
        		A[i] += p[k]*A[i+k];
        		B[i] += p[k]*B[i+k];
        	}
        }
        printf("%.12f\n", B[0]/(1.0-A[0]));
	}
	return 0;
}


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zoj 3329 One Person Game(有环的概率dp)

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=3754 开始看错题意了,以为没翻到a,b,c时要在原来的基础上加a+b+c,按我的意思推出来一个公式,没想到样例还过了,简直无法debug。 公式很好推,设dp[i]表示当前为i分时到达目标状态需要投掷的期望,可转移到两个状态dp[0]和dp[i+k]。设转移

1168

ZOJ3329 One Person Game(待定系数法求期望)

ZOJ3329 原来成环可以不用高斯消元,待定系数法也可以 定义dp[i]dp[i]dp[i]是当前分数为iii,到达目标的期望次数 dp[i]=1+dp[0]∗p[0]+∑k=i+3i+k1+k2+k3dp[k]∗p[k]dp[i]=1+dp[0]*p[0]+\sum\limits_{k=i+3}^{i+k1+k2+k3}dp[k]*p[k]dp[i]=1+dp[0]∗p[0]+k=i+3∑i+k1+k2+k3​dp[k]∗p[k] 其中p[i]p[i]p[i]为每个分数出现的概率,这个可以预处理出来 发

jziwjxjd的博客 495

ZOJ 3329 One Person Game [概率DP]

There is a very simple and interesting one-person game. 给出3个骰子,每个骰子的面数已知,并且对于每个骰子来说每一面出现的概率是相等的,当三个骰子分别为A,B,C时,计数器清0,否则加上三个骰子的点数和,问计数器大于N的期望投掷数。(提取求解项的系数,化解公式)

Asuna_acmer 509

[ZOJ3329]One Person Game(概率dp)

题目描述传送门题解令f(i)表示得分为i时到结束的期望 显然这道题>0的点都可以从0转移过来,0也可以从每一个>0的点转移过来 所以这题就应该画柿子了 f(i)=∑f(i+k)p(k)+f(0)P+1f(i)=\sum f(i+k)p(k)+f(0)P+1,其中p(i)表示掷出和为i的骰子的概率,P表示掷出a,b,c的概率 每一项都包含f(0),令f(i)=pa(i)f(0)+pb(i),那

zyf2000 575

ZOJ3329 One Person Game 概率dp

/* 题目描述:有三个骰子,每个上面的数值分别是1~K1 , 1 ~K2 , 1 ~K3,有一个计数器,其初始值为0,现在抛三枚 骰子,如果当前计数器的值加上这三枚骰子的点数和大于n,那么游戏结束,否则将计数器的值加上点数和后 游戏继续,但是,如果掷出的三枚骰子的值为a,b,c时,计数器的值清零,先问游戏中投掷骰子的期望是多少? 思路:设dp[

jijijix的博客 369

ZOJ 3329 One Person Game (概率DP)

题目: LINK

Napoleon 829

概率dp练习 (16.04.25)

继续练习之旅,记录3道概率dp的题。希望自己从中收获并成长。zoj 3822 Dominationhttp://www.icpc.moe/onlinejudge/showProblem.do?problemCode=3822 大意:一天放一个棋子到棋盘中,最后要求每一行每一列至少一个棋子。求解天数期望。 当一个棋子放在一个位置 (i,j)时,所在的行和列均会收到影响。 根据这一点,列出转移方程

theArcticOcean 859
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