my @a=(15,24,23,110,30,58,3,22,79,6);
my ($x, $cnt) = 0;
my $wantedSentence= 250;

map {$x += $_; ++$cnt unless($x >= $wantedSentence)}@a;
print "This sentence in chapter " . ++$cnt;
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use strict;
use warnings;

my @chapters          = ( 15, 24, 23, 110, 30, 58, 3, 22, 79, 6 );
my $base              = 1;
my %sentenceToChapter =
   map {
           my $chapter = $_;
           my @range   = ( $base .. $base + $chapters[ $chapter ] - 1 
+);
           $base       += scalar @range;
           map { $_ => $chapter + 1 } @range;
       }
   0 .. $#chapters;

print qq{Sentence $_ is in chapter $sentenceToChapter{ $_ }\n}
   for qw{ 250 15 16 98 };
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The output.

Sentence 250 is in chapter 6
Sentence 15 is in chapter 1
Sentence 16 is in chapter 2
Sentence 98 is in chapter 4
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I hope this is useful.

Cheers,

JohnGG

construct a lookup hash mapping sentences to chapters

The keys of this hash turn out to be the consecutive numbers from one up to the total number of sentences. The lookup could just as well be an array.

my @chapters          = ( 15, 24, 23, 110, 30, 58, 3, 22, 79, 6 );
my $chapter = 1;
my @sentenceToChapter = (
    undef, # start counting at 1
    map { ($chapter++) x $_ } @chapters,
);

print qq{Sentence $_ is in chapter $sentenceToChapter[ $_ ]\n}
   for qw{ 250 15 16 98 };
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my @a=(15,24,23,110,30,58,3,22,79,6);
my $wantedSentence = 250;
my $chptr = 1;
foreach (@a) {
   $wantedSentence -= $_;
   if ($wantedSentence <= 0) {
       print "$chptr";
       last;
   }
   $chptr++;
}
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In essence, you are trying to determine a hidden property of a number (which chapter a sentence is in) by looking at some other property of the number (its ordinal value in all sentences). Are you looking for a way to create a function that given any sentence in the book, determine which chapter it is in by some creative math approach? If so, please respond. If not, probably the most efficient way is to map all sentences to a chapter in the book using a hash (assumes book is static, numerous lookups, and that memory consumption is not prohibitive).

In case we're voting, I agree with Anonymous Monk that I would be interested in a creative-math approach to finding the chapter of a given book containing a given sentence.

JadeNB,
In a nutshell, this problem boils down to where this number fits in a list. The obvious non-creative solution to this is a binary search because it is both space and time efficient.

It would also be fairly trivial if each chapter had the same number of sentences. One trick might be then to find the chapter with the most number of sentences and pad each chapter to be equal to that chapter. When you "hand out" sentences and their ordinal value, there will be gaps representing the padding but you can then instantly take any sentence and find which chapter it belongs to. You can also find its true overall ordinal value in the book as well as the ordinal value in the chapter. Consider the following:

Chapter 1:  3 sentences
Chapter 2:  5 sentences
Chapter 3:  4 sentences
Chapter 4:  6 sentences

The chapter with the longest chapter is 4, so we pad all of them to 6

Chapter 1:  01, 02, 03             # 04, 05, 06 are padded
     real:  01, 02, 03
Chapter 2:  07, 08, 09, 10, 11     # 12 is padded
     real:  04, 05, 06, 07, 08
Chapter 3:  13, 14, 15, 16         # 17 and 18 are padded
     real:  09, 10, 11, 12
Chapter 4:  19, 20, 21, 22, 23, 24 # no padding
     real:  13, 14, 15, 16, 17, 18

Now let's say someone says, what chapter does 15 belong to?

ceil(15 / 6) = 3

Now let's say someone wants to know the true ordinal overall value

my %pad = (1 => 0, 2 => 3, 3 => 4, 4 => 6);

15 - $pad{ceil(15 / 6)} = 11

Now let's say someone wants to know what is the ordinal value of the s
+entence within the chapter

15 mod 6 = 3
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Cheers - L~R