One Person Game
Time Limit: 1 Second Memory Limit: 32768 KB Special Judge
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
2 0 2 2 2 1 1 1 0 6 6 6 1 1 1
Sample Output
1.142857142857143 1.004651162790698
题目大意:给出了k1,k2,k3三个筛子。当k1 == a k2 == b k3 == c时分数归零,否则累加,问当总和到n以上须要的次数期望
状态方程非常好写。dp[i]代表由i到n以上须要的次数,dp[i] = ∑(p[j]*dp[i+j])+q*dp[0] + 1。p[j]代表掷出和为j的概率,q为归零的概率。可是为问题出现了,在状态方程中有dp[0]这是我们要求解的值。所以要带入系数dpa[],dpb[],dp[i] = dpa[i] + dpb[i]*dp[0] ;
最后求解出dp[0] = dpa[0] + dpb[0]*dp[0],能够解除dp[0];
dp[i] = dpa[i] + dpb[i]*dp[0] = ∑(p[j]*dp[i+j])+ q*dp[0]+1;
得到dpa[i] = ∑( p[j]*dpa[i+j] ) + 1 ; dpb[i] = ∑( p[j]*dpb[i+j] ) + q ;
#include <cstdio> #include <cstring> #include <algorithm> using namespace std; double p[20] , q , cnt ; double dpa[600] , dpb[600] ; int main() { int t , n , m , i , j , k , l , k1 , k2 , k3 , a , b , c ; scanf("%d", &t); while(t--) { scanf("%d %d %d %d %d %d %d", &n, &k1, &k2, &k3, &a, &b, &c); memset(p,0,sizeof(p)); memset(dpa,0,sizeof(dpa)); memset(dpb,0,sizeof(dpb)); p[a+b+c] = -1 ; cnt = 0 ; m = k1 + k2 + k3 ; for(i = 1 ; i <= k1 ; i++) for(j = 1 ; j <= k2 ; j++) for(k = 1 ; k <= k3 ; k++) { p[i+j+k] += 1.0 ; cnt += 1.0 ; } for(i = 3; i <= m ; i++) p[i] /= cnt ; q = 1.0 / cnt ; for(i = n ; i >= 0 ; i--) { dpa[i] = 1.0 ; dpb[i] = q ; for(j = 3 ; j <= k1+k2+k3 ; j++) { dpa[i] += p[j]*dpa[i+j] ; dpb[i] += p[j]*dpb[i+j] ; } } printf("%.10lf\n", dpa[0]/(1-dpb[0])); } return 0; }
时间: 2024-10-25 12:26:25