给定一个函数,它产生的是1到5之间的随机整数,写一个函数,它产生的是1到7之间的随机整数。
当前回答
我不喜欢从1开始的范围,所以我将从0开始:-)
unsigned rand5()
{
return rand() % 5;
}
unsigned rand7()
{
int r;
do
{
r = rand5();
r = r * 5 + rand5();
r = r * 5 + rand5();
r = r * 5 + rand5();
r = r * 5 + rand5();
r = r * 5 + rand5();
} while (r > 15623);
return r / 2232;
}
其他回答
这类似于@RobMcAfee,除了我使用魔术数字而不是2维数组。
int rand7() {
int m = 1203068;
int r = (m >> (rand5() - 1) * 5 + rand5() - 1) & 7;
return (r > 0) ? r : rand7();
}
int rand7()
{
int zero_one_or_two = ( rand5() + rand5() - 1 ) % 3 ;
return rand5() + zero_one_or_two ;
}
Here's a solution that fits entirely within integers and is within about 4% of optimal (i.e. uses 1.26 random numbers in {0..4} for every one in {0..6}). The code's in Scala, but the math should be reasonably clear in any language: you take advantage of the fact that 7^9 + 7^8 is very close to 5^11. So you pick an 11 digit number in base 5, and then interpret it as a 9 digit number in base 7 if it's in range (giving 9 base 7 numbers), or as an 8 digit number if it's over the 9 digit number, etc.:
abstract class RNG {
def apply(): Int
}
class Random5 extends RNG {
val rng = new scala.util.Random
var count = 0
def apply() = { count += 1 ; rng.nextInt(5) }
}
class FiveSevener(five: RNG) {
val sevens = new Array[Int](9)
var nsevens = 0
val to9 = 40353607;
val to8 = 5764801;
val to7 = 823543;
def loadSevens(value: Int, count: Int) {
nsevens = 0;
var remaining = value;
while (nsevens < count) {
sevens(nsevens) = remaining % 7
remaining /= 7
nsevens += 1
}
}
def loadSevens {
var fivepow11 = 0;
var i=0
while (i<11) { i+=1 ; fivepow11 = five() + fivepow11*5 }
if (fivepow11 < to9) { loadSevens(fivepow11 , 9) ; return }
fivepow11 -= to9
if (fivepow11 < to8) { loadSevens(fivepow11 , 8) ; return }
fivepow11 -= to8
if (fivepow11 < 3*to7) loadSevens(fivepow11 % to7 , 7)
else loadSevens
}
def apply() = {
if (nsevens==0) loadSevens
nsevens -= 1
sevens(nsevens)
}
}
如果你将一个测试粘贴到解释器中(实际上是REPL),你会得到:
scala> val five = new Random5
five: Random5 = Random5@e9c592
scala> val seven = new FiveSevener(five)
seven: FiveSevener = FiveSevener@143c423
scala> val counts = new Array[Int](7)
counts: Array[Int] = Array(0, 0, 0, 0, 0, 0, 0)
scala> var i=0 ; while (i < 100000000) { counts( seven() ) += 1 ; i += 1 }
i: Int = 100000000
scala> counts
res0: Array[Int] = Array(14280662, 14293012, 14281286, 14284836, 14287188,
14289332, 14283684)
scala> five.count
res1: Int = 125902876
分布很好,很平坦(在每个箱子中,10^8的1/7大约在10k范围内,就像预期的近似高斯分布一样)。
这里允许作业题吗?
这个函数进行粗略的“以5为基数”的数学运算,生成0到6之间的数字。
function rnd7() {
do {
r1 = rnd5() - 1;
do {
r2=rnd5() - 1;
} while (r2 > 1);
result = r2 * 5 + r1;
} while (result > 6);
return result + 1;
}
PHP解决方案
<?php
function random_5(){
return rand(1,5);
}
function random_7(){
$total = 0;
for($i=0;$i<7;$i++){
$total += random_5();
}
return ($total%7)+1;
}
echo random_7();
?>