loj 1030概率dp

题目链接：http://lightoj.com/volume_showproblem.php?problem=1030

思路：一直以来对这种概率题都挺感冒的=.=......还是说一下思路吧，dp[i]表示前i个位置所能能到的期望值，然后我们可以从后往前递推。

 1 #include <iostream>

 2 #include <cstdio>

 3 #include <cstring>

 4 #include <algorithm>

 5 #include <cmath>

 6 using namespace std;

 7 

 8 const int MAXN = (100 + 10);

 9 double dp[MAXN], num[MAXN];

10 int n;

11 

12 int main()

13 {

14     int _case, t = 1;

15     scanf("%d", &_case);

16     while (_case--) {

17         scanf("%d", &n);

18         for (int i = 1; i <= n; i++) {

19             scanf("%lf", &num[i]);

20         }

21         memset(dp, 0, sizeof(dp));

22         dp[n] = num[n];

23         for (int i = n - 1; i >= 1; i--) {

24             int d = min(6, n-i);

25             dp[i] = num[i];

26             for (int j = 1; j <= d; j++) {

27                 dp[i] += dp[i+j]/d;

28             }

29         }

30         printf("Case %d: %.7lf\n", t++, dp[1]);

31     }

32     return 0;

33 }

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