A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 99.
Find the largest palindrome made from the product of two 3-digit numbers.
一个回文数指的是从左向右和从右向左读都一样的数字。最大的由两个两位数乘积构成的回文数是9009 = 91 * 99.
找出最大的有由个三位数乘积构成的回文数。
这个题目难度不大,写一个小程序就OK了,
思路分为两步:
第一:因为999*999=998001,所以从998001开始向前,找回文数,因为我们要的是最大的,所以从大往小找
第二:找到以后验证是否是两个三位数的乘积,这里的验证也是从大往小,看看能不能被999整除,如果能,检查商是不是三位数,如果是,那么结果就出来了。
一下为程序:
use strict; use warnings; my $i; my $flag; my $num; for($i=998001;$i>100000;$i--) { $num=$i; my $six=(int($num/100000)); my $five=(int($num/10000)-10*int($num/100000)); my $four=(int($num/1000)-10*int($num/10000)); my $three=(int($num/100)-10*int($num/1000)); my $two=(int($num/10)-10*int($num/100)); my $one=($num-10*int($num/10)); if($six==$one && $five==$two && $four==$three) { my $cout=999; for($cout=999;$cout>100;$cout--) { if(0==$num%$cout) { my $another=$num/$cout; if($another>100 && $another<999) { print "$num = $cout X $another \n"; $flag=1; last; } } } last if $flag==1; } }
C:\WINDOWS\system32\cmd.exe /c perl "F:\perl\c.pl 906609 = 993 X 913 Hit any key to close this window...
题目如下:
2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.
What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20?
2520是最小的能被1-10中每个数字整除的正整数。
最小的能被1-20中每个数整除的正整数是多少?
这里的第二题其实都不用谢程序的,直接拿个计算器算一下就好了,
如下:
$i=2*3*2*5*7*2*3*11*13*2*17*19; print $i."\n";
C:\WINDOWS\system32\cmd.exe /c perl "F:\perl\c.pl" 232792560 Hit any key to close this window...
题目如下:
The sum of the squares of the first ten natural numbers is,
The square of the sum of the first ten natural numbers is,
Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is 3025 385 = 2640.(和的平方-平方的和=?)
Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.
求1到100 和的平方-平方的和=?
程序如下:
use strict; use warnings; my $sum=0; my $sum_square=0; my $square_sum=0; my $cout; my $difference; for($cout=1;$cout<101;$cout++) { $square_sum=$square_sum+$cout*$cout; $sum=$sum+$cout; } $sum_square=$sum*$sum; $difference=$square_sum-$sum_square; print "$difference = $square_sum-$sum_square\n";
C:\WINDOWS\system32\cmd.exe /c perl "F:\perl -25164150 = 338350-25502500 Hit any key to close this window...
题目如下:
By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6th prime is 13.
What is the 10 001st prime number?
也就是着第10001个素数:程序如下:
use strict; use warnings; my $cout=1; my $flag=0; my $i; for($i=3;;$i+=2)#这里是间隔2,这样可以节省一半的时间 { foreach(2..$i**0.5) { if(0==$i%$_) { $flag=1; last; } else { next; } } $cout=$cout+1 if($flag==0); $flag=0; if($cout==10001) { print "$i\n"; last; } }
结果如下:
C:\WINDOWS\system32\cmd.exe /c perl "F:\perl\c.pl" 104743 Hit any key to close this window...
这个题目非常的长:
如下:
找出以下这个1000位的整数中连续5个数字的最大乘积。(例如前五个数字的乘积是7*3*1*6*7=882)
73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450
我们把所有的数放在一个数组里,然后从第一个元素开始依次相乘,并将得到的结果进行比较,大的留下。
这里要说的把1000个数中间都加一个空格,然后变成数组,用vim的宏就很简单的完成了。
程序如下:
my $big=0; my $sum=0; my @num=qw/7 3 1 6 7 1 7 6 5 3 1 3 3 0 6 2 4 9 1 9 2 2 5 1 1 9 6 7 4 4 2 6 5 7 4 7 4 2 3 5 5 3 4 9 1 9 4 9 3 4 9 6 9 8 3 5 2 0 3 1 2 7 7 4 5 0 6 3 2 6 2 3 9 5 7 8 3 1 8 0 1 6 9 8 4 8 0 1 8 6 9 4 7 8 8 5 1 8 4 3 8 5 8 6 1 5 6 0 7 8 9 1 1 2 9 4 9 4 9 5 4 5 9 5 0 1 7 3 7 9 5 8 3 3 1 9 5 2 8 5 3 2 0 8 8 0 5 5 1 1 1 2 5 4 0 6 9 8 7 4 7 1 5 8 5 2 3 8 6 3 0 5 0 7 1 5 6 9 3 2 9 0 9 6 3 2 9 5 2 2 7 4 4 3 0 4 3 5 5 7 6 6 8 9 6 6 4 8 9 5 0 4 4 5 2 4 4 5 2 3 1 6 1 7 3 1 8 5 6 4 0 3 0 9 8 7 1 1 1 2 1 7 2 2 3 8 3 1 1 3 6 2 2 2 9 8 9 3 4 2 3 3 8 0 3 0 8 1 3 5 3 3 6 2 7 6 6 1 4 2 8 2 8 0 6 4 4 4 4 8 6 6 4 5 2 3 8 7 4 9 3 0 3 5 8 9 0 7 2 9 6 2 9 0 4 9 1 5 6 0 4 4 0 7 7 2 3 9 0 7 1 3 8 1 0 5 1 5 8 5 9 3 0 7 9 6 0 8 6 6 7 0 1 7 2 4 2 7 1 2 1 8 8 3 9 9 8 7 9 7 9 0 8 7 9 2 2 7 4 9 2 1 9 0 1 6 9 9 7 2 0 8 8 8 0 9 3 7 7 6 6 5 7 2 7 3 3 3 0 0 1 0 5 3 3 6 7 8 8 1 2 2 0 2 3 5 4 2 1 8 0 9 7 5 1 2 5 4 5 4 0 5 9 4 7 5 2 2 4 3 5 2 5 8 4 9 0 7 7 1 1 6 7 0 5 5 6 0 1 3 6 0 4 8 3 9 5 8 6 4 4 6 7 0 6 3 2 4 4 1 5 7 2 2 1 5 5 3 9 7 5 3 6 9 7 8 1 7 9 7 7 8 4 6 1 7 4 0 6 4 9 5 5 1 4 9 2 9 0 8 6 2 5 6 9 3 2 1 9 7 8 4 6 8 6 2 2 4 8 2 8 3 9 7 2 2 4 1 3 7 5 6 5 7 0 5 6 0 5 7 4 9 0 2 6 1 4 0 7 9 7 2 9 6 8 6 5 2 4 1 4 5 3 5 1 0 0 4 7 4 8 2 1 6 6 3 7 0 4 8 4 4 0 3 1 9 9 8 9 0 0 0 8 8 9 5 2 4 3 4 5 0 6 5 8 5 4 1 2 2 7 5 8 8 6 6 6 8 8 1 1 6 4 2 7 1 7 1 4 7 9 9 2 4 4 4 2 9 2 8 2 3 0 8 6 3 4 6 5 6 7 4 8 1 3 9 1 9 1 2 3 1 6 2 8 2 4 5 8 6 1 7 8 6 6 4 5 8 3 5 9 1 2 4 5 6 6 5 2 9 4 7 6 5 4 5 6 8 2 8 4 8 9 1 2 8 8 3 1 4 2 6 0 7 6 9 0 0 4 2 2 4 2 1 9 0 2 2 6 7 1 0 5 5 6 2 6 3 2 1 1 1 1 1 0 9 3 7 0 5 4 4 2 1 7 5 0 6 9 4 1 6 5 8 9 6 0 4 0 8 0 7 1 9 8 4 0 3 8 5 0 9 6 2 4 5 5 4 4 4 3 6 2 9 8 1 2 3 0 9 8 7 8 7 9 9 2 7 2 4 4 2 8 4 9 0 9 1 8 8 8 4 5 8 0 1 5 6 1 6 6 0 9 7 9 1 9 1 3 3 8 7 5 4 9 9 2 0 0 5 2 4 0 6 3 6 8 9 9 1 2 5 6 0 7 1 7 6 0 6 0 5 8 8 6 1 1 6 4 6 7 1 0 9 4 0 5 0 7 7 5 4 1 0 0 2 2 5 6 9 8 3 1 5 5 2 0 0 0 5 5 9 3 5 7 2 9 7 2 5 7 1 6 3 6 2 6 9 5 6 1 8 8 2 6 7 0 4 2 8 2 5 2 4 8 3 6 0 0 8 2 3 2 5 7 5 3 0 4 2 0 7 5 2 9 6 3 4 5 0/; for($i=0;$i<996;$i++) { $sum=$num[$i]*$num[$i+1]*$num[$i+2]*$num[$i+3]*$num[$i+4]; if($sum>$big) { $big=$sum; } else { next; } } print $big."\n";结果如下:
C:\WINDOWS\system32\cmd.exe /c perl "F:\perl\c.pl" 40824 Hit any key to close this window...
A Pythagorean triplet is a set of three natural numbers, a b c, for which,
For example, 32 + 42 = 9 + 16 = 25 = 52.
There exists exactly one Pythagorean triplet for which a + b + c = 1000.
Find the product abc.
一个毕达哥拉斯三元组是一个包含三个自然数的集合,a<b<c,满足条件:
例如:32 + 42 = 9 + 16 = 25 = 52.
已知存在并且只存在一个毕达哥拉斯三元组满足条件a + b + c = 1000。
找出该三元组中abc的乘积。
程序如下:
use strict; use warnings; my $first; my $second; my $sum; for($first=1;$first<333;$first++) { for($second=$first+1;$second<500;$second++) { if($first**2+$second**2==(1000-$first-$second)**2) { print "$first X $second X(1000-$first-$second)="; $sum=$first*$second*(1000-$first-$second); print $sum."\n"; last; } else { next; } } }
C:\WINDOWS\system32\cmd.exe /c perl "F:\perl\c.pl" 200 X 375 X(1000-200-375)=31875000 Hit any key to close this window...
这个题目和前面求素数的问题类似,具体的方案就不说了,直接帖代码:
use strict; use warnings; my $i; my $sum=2; my $cout; my $flag=0; for($i=3;$i<2000000;$i+=2) { $flag=0; for($cout=2;$cout<$i**0.5+1;$cout++) { if(0==$i%$cout) { $flag=1; last; } } $sum=$sum+$i if($flag==0); } print $sum."\n";
C:\WINDOWS\system32\cmd.exe /c perl "F:\perl\c.pl" 142913828922 Hit any key to close this window...