题目
{% fold 点击显/隐题目 %}
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.
题解
题目思维比较独特,需要一些数学技巧
如图,设 dp[i]
表示当前计数为 i
的时候结束游戏的期望值
则在状态 i
的时候可以转移到状态 0
和 状态 i+j
计算出当前骰子投出和为 j
的概率为 p[j]
(dp[0]是投出a
,b
,c
的概率)
则有 dp[i] = dp[0]p[0] + sum{ dp[i+j]p[j] }
然后会发现一个很重要的问题,状态同时向两侧转移
也即无论是从大到小还是从小到大或者其他转移形式,这个式子都不满足无后效性
但是这里可以看出,往小了转移只有 dp[0]
一个状态
可以用数学重新推导一下
{% raw %}$
\begin{align*}
dp[i] &= \sum (dp[i+j] \times p[j]) + dp[0] \times p[0] \
设: dp[i] &= A[i] \times dp[0] + B[i] \
带入原式得: dp[i] &= \sum((A[i+j] \times dp[0] + B[i+j) \times p[j]) + dp[0] \times p[0] \
整理得: dp[i] &=\sum ((p[j]*A[i+j]) + p[0]) \times dp[0]) + \sum(p[j] \times B[i+j]) + 1 \
\
\therefore A[i] &= \sum(p[j] \times A[i+j]) + p[0] \
B[i] &= \sum (p[j] \times B[i+j]) + 1 \
\
且: dp[0] &= \frac{B[0]}{1-A[0]} \
A[i] &= B[i] = 0,i>n
\end{align*}
${% endraw %}
这样只需要预先处理出 p[i]
然后用 O(n)
的时间求出 A[i]
,B[i]
即可
最后结果就是 B[0]/(1-A[0])
代码
{% fold 点击显/隐代码 %}```cpp One Person Game https://github.com/OhYee/sourcecode/tree/master/ACM 代码备份
#include
#include
using namespace std;
#define Log(format, ...) //printf(format, ##VA_ARGS)
const int maxk = 3;
const int maxnum = 20;
const int maxn = 550;
int n, k[maxk], a[maxk];
double A[maxn], B[maxn], p[maxnum];
int main() {
int T;
scanf("%d", &T);
while (T--) {
scanf("%d%d%d%d%d%d%d", &n, &k[0], &k[1], &k[2], &a[0], &a[1], &a[2]);
memset(p, 0, sizeof(p));
for (int i = 1; i <= k[0]; ++i)
for (int j = 1; j <= k[1]; ++j)
for (int l = 1; l <= k[2]; ++l) {
if (i == a[0] && j == a[1] && l == a[2]) {
p[0] += 1;
} else {
p[i + j + l] += 1;
}
}
int sum = k[0] * k[1] * k[2];
for (int i = 0; i < maxnum; ++i) {
p[i] = p[i] / sum;
Log("%d %f\n", i, p[i]);
}
memset(A, 0, sizeof(A));
memset(B, 0, sizeof(B));
for (int i = n; i >= 0; --i) {
for (int j = 3; j < maxnum; ++j) {
Log("%d %d %f %f %f\n", i,j, p[j],A[i+j],B[i+j]);
A[i] += p[j] * A[i + j];
B[i] += p[j] * B[i + j];
}
Log("%d %f %f\n", i, A[i],B[i]);
A[i] += p[0];
B[i] += 1;
Log("%d %f %f\n", i, A[i],B[i]);
}
printf("%.15f\n", B[0] / (1 - A[0]));
}
return 0;
}
{% endfold %}