如何平滑隨機分佈？

Question

我試圖隨機字符的出現在遊戲中，但用自己的名字作爲種子。所以，如果你以前遇到過「鮑勃」在比賽中，他永遠有相同的頭髮/眼睛等

現在我只是從生成的名稱中的數字（加入所有的字符代碼）和然後使用模數來決定他們擁有哪些選項。

實施例：Bob的種子是276（66 + 111 + 98）。 276％的髮型（40）的數量在36

這工作正常，但爲350+人的名單產生，分佈如下：

hair style: 0/# of people using it: 15 
hair style: 1/# of people using it: 8 
hair style: 2/# of people using it: 4 
hair style: 3/# of people using it: 5 
hair style: 4/# of people using it: 7 
hair style: 5/# of people using it: 5 
hair style: 6/# of people using it: 7 
hair style: 7/# of people using it: 14 
hair style: 8/# of people using it: 12 
hair style: 9/# of people using it: 6 
hair style: 10/# of people using it: 7 
hair style: 11/# of people using it: 2 
hair style: 12/# of people using it: 7 
hair style: 13/# of people using it: 10 
hair style: 14/# of people using it: 11 
hair style: 15/# of people using it: 7 
hair style: 16/# of people using it: 12 
hair style: 17/# of people using it: 7 
hair style: 18/# of people using it: 6 
hair style: 19/# of people using it: 10 
hair style: 20/# of people using it: 5 
hair style: 21/# of people using it: 10 
hair style: 22/# of people using it: 11 
hair style: 23/# of people using it: 3 
hair style: 24/# of people using it: 6 
hair style: 25/# of people using it: 8 
hair style: 26/# of people using it: 5 
hair style: 27/# of people using it: 11 
hair style: 28/# of people using it: 10 
hair style: 29/# of people using it: 6 
hair style: 30/# of people using it: 13 
hair style: 31/# of people using it: 11 
hair style: 32/# of people using it: 10 
hair style: 33/# of people using it: 12 
hair style: 34/# of people using it: 3 
hair style: 35/# of people using it: 11 
hair style: 36/# of people using it: 9 
hair style: 37/# of people using it: 4 
hair style: 38/# of people using it: 10 
hair style: 39/# of people using it: 15 

的分佈不是很順暢，它遍佈全球（不出所料）。我會碰到很多髮型＃0的人，並且沒有髮型＃11的人。

我怎樣才能順利完成這一出位？

Answer 1

，如果你使用真正的哈希函數，而不是僅僅求和ASCII/Unicode碼點你可能有更好的運氣。 The djb2 function頗爲流行的ASCII輸入：

unsigned long hash(unsigned char *str) { 
    unsigned long hash = 5381; 
    int c; 

    while (c = *str++) 
     hash = ((hash << 5) + hash) + c; /* hash * 33 + c */ 

    return hash; 
} 

因此，例如，

unsigned long hairStyle = hash(characterName) % kHairStyleCount; 

參見What's a good hash function for English words?