题目转换为,对数p,初值为2007,当p为奇数时,加一,为偶数时,除二.
用Perl语言写一程序求解,答案为:
第12次操作时,x的指数为4
第14次操作时,x的指数为1
继续下去出现循环,即不停的1,2,1,2,1,2,1,2,...
程序如下供你参考:
$p = 2007; # $p为指数
$n = 0; # $n为操作次数
do
{
$n ++;
if (($p % 2) == 0)
{
$p /= 2;
}
else
{
$p += 1;
}
if ($p == 4)
{
print "$n:This time,p = 4n";
}
if ($n == 14)
{
print "When 14th time,p = $pn";
}
if ($p == 1)
{
print "When $n time,p = 1n";
$w = ;
return;
}
} while 1;
程序输出为:
12:This time,p = 4
When 14th time,p = 1
When 14 time,p = 1