Light OJ 1030-Discovering Gold (expected)
1030-Discovering Gold
|
PDF (English) |
Statistics |
Forum |
| Time Limit: 2 second (s) |
Memory Limit: 32 MB |
You are in a cave, a long cave! The cave can be represented by1 x NGrid. Each cell of the cave can contain any amount of gold.
Initially you are in position1. Now each turn you throw a perfect6Sided dice. If you getXIn the dice after throwing, you addXTo your position and collect all the gold from the new position. If your new position is outside the cave, then you keep throwing again until you get a suitable result. When you reachNthPosition you stop your journey. Now you are given the information about the cave, you have to find outExpectedNumber of gold you can collect using the given procedure.
Input
Input starts with an integerT (≤ 100), Denoting the number of test cases.
Each case contains a blank line and an integerN (1 ≤ N ≤ 100)Denoting the dimension of the cave. The next line containsNSpace separated integers.IthInteger of this line denotes the amount of gold you will get if you come toIthCell. You may safely assume that all the given integers will be non-negative and no integer will be greater1000.
Output
For each case, print the case number and the expected number of gold you will collect. Errors less10-6Will be ignored.
| Sample Input |
Output for Sample Input |
3 1 101 2 10 3 3 3 6 9 |
Case 1: 101.0000000000 Case 2: 13.000 Case 3: 15 |
# Include
# Include
# Include
# Include
# Include
# Include
# Include
# Include
# Include
# Define L (x) (x <1) # define R (x) (x <1 | 1) # define MID (x, y) (x + y)> 1) # define eps 1e-8typedef _ int64 ll; # define fre (I, a, B) for (I = a; I
= A; I --) # define mem (t, v) memset (t), v, sizeof (t) # define ssf (n) scanf ("% s ", n) # define sf (n) scanf ("% d", & n) # define sff (a, B) scanf ("% d", &, & B) # define sfff (a, B, c) scanf ("% d", & a, & B, & c) # define pf printf # define bug pf ("Hi \ n") using namespace std; # define INF 0x3f3f3f3f # define N 10005 double dp [N]; int main () {int I, j, t, n, ca = 0; sf (t); while (t --) {mem (dp, 0); sf (n ); fre (I, 1, n + 1) scanf ("% lf", & dp [I]); for (I = n-1; I> = 1; I --) for (j = 1; j <= 6; j ++) dp [I] + = dp [I + j]/min (6, n-I ); // This is important, because if n is exceeded, it is invalid pf ("Case % d: %. 6lf \ n ", ++ ca, dp [1]);} return 0 ;}