ZOJ 3329 One Person Game——(概率DP+待定系数求解方程)
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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:
- Set the counter to 0 at first.
- 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.
- 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
20 2 2 2 1 1 10 6 6 6 1 1 1
Sample Output
1.1428571428571431.004651162790698
题目大意:
有三个骰子,分别有
(1) 首先将计数器设置为
(2) 如果骰子
(3) 如果计数器的值
现在计算游戏结束前掷骰子的次数期望。
解题思路:
然后观察这个式子发现,这个石子都跟
将式子
将这个式子与第
然后通过这两个递推式求出
解得
代码:
#include <iostream>#include <string.h>#include <string>#include <algorithm>#include <stdio.h>#include <stdlib.h>#include <math.h>#include <map>using namespace std;typedef long long LL;const int MAXN = 5e2+20;const double PI = acos(-1);const double eps = 1e-8;const LL MOD = 1e9+7;double p[20], A[MAXN], B[MAXN];void Init(){ memset(p, 0, sizeof(p)); memset(A, 0, sizeof(A)); memset(B, 0, sizeof(B));}int main(){ //freopen("C:/Users/yaonie/Desktop/in.txt", "r", stdin); //freopen("C:/Users/yaonie/Desktop/out.txt", "w", stdout); int T; while(~scanf("%d", &T)){ while(T--){ int n, k1, k2, k3, a, b, c; scanf("%d%d%d%d%d%d%d",&n,&k1,&k2,&k3,&a,&b,&c); Init(); p[0] = 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) continue; p[i+j+k] += p[0]; } } } for(int i=n; i>=0; i--){ A[i] = p[0], 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]); } } printf("%.15f\n",B[0]/(1-A[0])); } } return 0;}
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