ZOJ3329 概率DP

One Person Game

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Time Limit: 1 Second      Memory Limit: 32768 KB      Special Judge

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There is a very simple and interesting one-person game. You have 3 dice, namely Die1, Die2 and Die3. Die1 has K₁ faces. Die2 has K₂ faces. Die3 has K₃ faces. All the dice are fair dice, so the probability of rolling each value, 1 to K₁, K₂, K₃ is exactly 1 / K₁, 1 / K₂ and 1 / K₃. You have a counter, and the game is played as follow:

1. Set the counter to 0 at first.
2. Roll the 3 dice simultaneously. If the up-facing number of Die1 is a, the up-facing number of Die2 is b and the up-facing number of Die3 is c, set the counter to 0. Otherwise, add the counter by the total value of the 3 up-facing numbers.
3. If the counter's number is still not greater than n, go to step 2. Otherwise the game is ended.

Calculate the expectation of the number of times that you cast dice before the end of the game.

Input

There are multiple test cases. The first line of input is an integer T (0 < T <= 300) indicating the number of test cases. Then T test cases follow. Each test case is a line contains 7 non-negative integers n, K₁, K₂, K₃, a, b, c (0 <= n <= 500, 1 < K₁, K₂, K₃ <= 6, 1 <= a <= K₁, 1 <= b <= K₂, 1 <= c <= K₃).

Output

For each test case, output the answer in a single line. A relative error of 1e-8 will be accepted.

Sample Input

2
0 2 2 2 1 1 1
0 6 6 6 1 1 1

Sample Output

1.142857142857143
1.004651162790698

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Author: CAO, Peng
Source: The 7th Zhejiang Provincial Collegiate Programming Contest

题意：有三个骰子，面值分别是k1，k2，k3。每次扔出的值之和加到ans上，问多少次才能ans>n；当然，当遇到k1=a,k2=b,k3=c时，ans=0;重新开始累加。

思路：

设dp[i]表示达到i分时到达目标状态的期望，pk为投掷k分的概率，p0为回到0的概率
则dp[i]=∑(pk*dp[i+k])+dp[0]*p0+1;
都和dp[0]有关系，而且dp[0]就是我们所求，为常数
设dp[i]=A[i]*dp[0]+B[i];
代入上述方程右边得到：
dp[i]=∑(pk*A[i+k]*dp[0]+pk*B[i+k])+dp[0]*p0+1
=(∑(pk*A[i+k])+p0)dp[0]+∑(pk*B[i+k])+1;
明显A[i]=(∑(pk*A[i+k])+p0) ，B[i]=∑(pk*B[i+k])+1
先递推求得A[0]和B[0]. 那么 dp[0]=B[0]/(1-A[0]);

#include"bits/stdc++.h"

#define db double
#define ll long long
#define vl vector<ll>
#define ci(x) scanf("%d",&x)
#define cd(x) scanf("%lf",&x)
#define cl(x) scanf("%lld",&x)
#define pi(x) printf("%d
",x)
#define pd(x) printf("%f
",x)
#define pl(x) printf("%lld
",x)
#define rep(i, n) for(int i=0;i<n;i++)
using namespace std;
const int N = 1e6 + 5;
const int mod = 1e9 + 7;
const int MOD = 998244353;
const db PI = acos(-1.0);
const db eps = 1e-10;
const ll INF = 0x3fffffffffffffff;

int t;
int n,k1,k2,k3,a,b,c;
db A[605],B[605];
db p[605];
int main() {
    ci(t);
    while (t--) {
        ci(n),ci(k1),ci(k2),ci(k3),ci(a),ci(b),ci(c);
        memset(A,0, sizeof(A));
        memset(B,0, sizeof(B));
        memset(p,0, sizeof(p));
        db p0=1.0/(k1*k2*k3);
        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;
            for(int k=3;k<=k1+k2+k3;k++) {
                A[i] += p[k] * A[i + k];
                B[i] += p[k] * B[i + k];
            }
        }
        db ans=B[0]/(1-A[0]);
        printf("%.10f
", ans);
    }
    return 0;
}