Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

In the pictures below, you can see several types of Dices. It is very easy to assume that they are used in many different types of games. In this problem, the face of the dices will be like those on the first figure, that means each will have some dots it each face, which will indicate the value of that particular face. But the looks of the dices will be like those on the second figure, that means each dice may have 2, 3, 4, 5 up to 50 different faces. A dice with p faces have face values 1, 2, 3 … p. When a dice is thrown all the faces are equally likely to come on top. For example an ordinary dice in figure-1 has face values 1, 2, 3, 4, 5, 6 and the probability that face with value 1 will be on top is 1/6. The same is the probability for the other faces coming on top.

Figure 1: Two conventional dices	Figure 2: Dices of different shape

The situation becomes complicated when you throw a dice more than once dice. For example when you throw a dice twice, for any arbitrary incident the summation of the top two faces can be within the range (inclusive) 2...12. But their probability is not equally likely or 1/12. The probability that the sum will be two is 1/36, but the probability that some will be three is 2/36 and so on. Given the description of a dice throw, that’s how many faces it have, how many throws will be made and value for the sum, you will have to find the probability of obtaining that sum.

The input will contain several lines of input. Each line contains three integers F (The number of faces of the dice thrown, 1<=F<=50), N (The number of throws and 0<=N<=50) and S (The summation value for which you to find out the probability, 0<S<=4000).

For each line of input you will have to produce one line of output which will be of the form a/b (Here b = F^N). Here a/b means the probability value. Note that we don’t need the floating point value of probability.