9596. Word processor
Bessie the cow is working on an essay for
her writing class. Since her handwriting is quite bad, she decides to type the
essay using a word processor.
The essay contains n words,
separated by spaces. Each word is between 1 and 15 characters long, inclusive,
and consists only of uppercase or lowercase letters. According to the
instructions for the assignment, the essay has to be
formatted in a very specific way: each line should contain no more than k
characters, not counting spaces. Fortunately, Bessie’s word processor can
handle this requirement, using the following strategy:
·
If
Bessie types a word, and that word can fit on the current line, put it on that
line.
·
Otherwise,
put the word on the next line and continue adding to that line.
Of course, consecutive words on the same
line should still be separated by a single space. There should be no space at
the end of any line.
Unfortunately, Bessie's word processor just
broke. Please help her format her essay properly!
Input. The first line contains two integers n (1 ≤ n ≤ 100) and k (1 ≤ k ≤ 80). The next line contains n words
separated by single spaces. No word will ever be larger than k characters,
the maximum number of characters on a line.
|
Sample
input |
Sample
output |
|
10 7 hello my name is
Bessie and this is my essay |
hello my name is Bessie and this is my essay |
greedy
Read
the input words sequentially. The current word is printed in the current line
if the total length of words printed in the line does not exceed k. If,
when adding a word to the current line, the length of the line becomes greater
than k, then the word is printed in a new line.
Read the input words into the array s.
char s[100];
Read the input
data.
scanf("%d %d", &n, &k);
The variable ptr
stores the current length of the output string.
ptr = 0;
Read n
words sequentially.
while (n--)
{
Read the next
word s and find its length len.
scanf("%s", s);
len = strlen(s);
If ptr + len
≤ k, then the word s can
be printed on the current line. The length of the output string will not exceed
k characters.
if (ptr + len
<= k)
{
ptr += len;
printf("%s
", s);
}
Otherwise, the
word s should be printed on a new line. Let the length ptr of the
new current line be equal to len.
else
{
ptr = len;
printf("\n%s
", s);
}
}
Read the input data.
cin >> n >> k;
The variable ptr stores the current length of the output string.
ptr = 0;
Read n words sequentially.
while (n--)
{
Read the next word s and find its length len.
cin >> s;
len = s.size();
If ptr + len ≤ k, then the word s can be printed on the current line. The
length of the output string will not exceed k characters.
if (ptr + len <= k)
{
ptr += len;
cout << s << " ";
}
Otherwise, the word s should be printed on a new line. Let the
length ptr of the new current line be equal to len.
else
{
ptr = len;
cout << endl << s << "
";
}
}
Read the input data.
n, k = map(int, input().split())
s = input().split()
The variable ptr stores the current length of the output string.
ptr = 0
Iterate through the words w in
the list s.
for w in s:
Find the length of the word w.
lw = len(w)
If ptr + lw ≤ k, then the word w can be
printed on the current line. The length of the output string will not exceed k
characters.
if ptr + lw <= k:
ptr += lw
print(w, end=" ")
Otherwise, the word w should
be printed on a new line. Let the length ptr of the new current line be
equal to lw.
else:
ptr = lw
print()
print(w, end=" ")