任务/分配
将空格插入文本并生成格式均匀（对齐）文本的方法。我编写了工作 Java 代码，但有必要用 Fortran 编写。
我编写的工作 Java 代码
public class ProgramingLanguagesTheoryHomeWork {
public static void main(String[] args) {
String inputString = "Lorem ipsum dolor sit amet, consectetur adipiscing elit. Duis et ante at eros euismod molestie eu varius ipsum. Aenean gravida pharetra magna quis auctor. Nullam est leo, dictum ac luctus quis, aliquet eu neque. Donec in nisl enim, et malesuada odio. Nulla scelerisque tortor id justo porttitor et vulputate urna pharetra. Pellentesque mollis condimentum elementum. Cras tempor, turpis hendrerit adipiscing lacinia, dolor metus egestas velit, venenatis vulputate leo eros elementum tellus. Vivamus id dui erat. Sed dictum consequat nulla, sed gravida eros mattis in. Suspendisse ac lorem lorem, ut molestie justo. Quisque suscipit velit ut odio euismod elementum. Vestibulum tempus.";
String outputString = format(inputString, 50, 10);
System.out.println(outputString);
}
private static String format(String text, int lineLength, int beginning_indent) {
String result = "";
String[] words = text.split(" ");
int i = 0;
while( i < words.length ) {
int characterCount = 0;
ArrayList<String> wordsForLine = new ArrayList<String>();
int substractForInterval = i == words.length -1 ? 0:1;
while( ( i < words.length ) && (characterCount <= lineLength - words[i].length() - substractForInterval) ) {
if( i == 0 ) {
String beginningIndentWord = "";
for(int k=0;k<beginning_indent;k++) {
beginningIndentWord += " ";
}
wordsForLine.add(beginningIndentWord);
characterCount += beginning_indent + 1;
}
wordsForLine.add(words[i]);
characterCount += words[i].length() + 1; //the one is for the interval after each word
i++;
}
characterCount --; //substract one character for the final interval which must not exist in the formatted text
int numberOfSpacesToAddAfterEachWord = (lineLength - characterCount) / (wordsForLine.size() -1); //substract one word becouse the final word cant really take an inteterval
int numberOfLeftOverSpaces = (lineLength - characterCount) % (wordsForLine.size() -1); //substract one word becouse the final word cant really take an inteterval
int numberOfLeftOverSpacesAdded = 0;
for(int j=0;j<wordsForLine.size();j++) {
result += wordsForLine.get(j);
if( j < (wordsForLine.size() - 1) ) //check to see if its the last word of the line and not add interval to it
result += " ";
if( i != words.length && j < (wordsForLine.size() - 1) ) { //check to see if its the last line that does not need to be converted and it its not the last character from a line which can not take intervals
for(int p=0;p<numberOfSpacesToAddAfterEachWord;p++) {
result += " ";
}
if(numberOfLeftOverSpacesAdded < numberOfLeftOverSpaces ) {
result += " ";
numberOfLeftOverSpacesAdded++;
}
}
}
result += "\n";
}
return result;
}
}
用 Fortran 重写
最初的问题
我只是无法弄清楚我将如何绕过可变长度字符串以及 List 变量。
当前问题
好的经过大量的努力，我终于推出了这个丑陋的代码。
- 但是，在将空格连接到字符串时，我似乎遗漏了一些关键点，或者至少我是这么认为的。
- 它没有输出它应该输出的东西。
program xfunc
implicit none
CHARACTER(LEN=682) :: text
CHARACTER(100) :: words(100)
CHARACTER(2000) :: result
integer :: lineLenght
integer :: beginningIndent
integer :: pos1
integer :: pos2
integer :: n
integer :: i
integer :: characterCount
CHARACTER(100) :: wordsForLine(100)
integer :: wordsForLinePointer
integer :: substractForInterval
CHARACTER(20) :: beginningIndentWord
integer :: k
integer :: numberOfSpacesToAddAfterEachWord
integer :: numberOfLeftOverSpaces
integer :: numberOfLeftOverSpacesAdded
integer :: j
integer :: p
text = 'Lorem ipsum dolor sit amet, &
consectetur adipiscing elit. Duis et &
ante at eros euismod molestie eu &
varius ipsum. Aenean gravida pharetra &
magna quis auctor. Nullam est leo, &
dictum ac luctus quis, aliquet eu &
neque. Donec in nisl enim, et &
malesuada odio. Nulla scelerisque &
tortor id justo porttitor et &
vulputate urna pharetra. Pellentesque &
mollis condimentum elementum. Cras &
tempor, turpis hendrerit adipiscing &
lacinia, dolor metus egestas velit, &
venenatis vulputate leo eros elementum &
tellus. Vivamus id dui erat. Sed &
dictum consequat nulla, sed gravida &
eros mattis in. Suspendisse ac lorem &
lorem, ut molestie justo. Quisque &
suscipit velit ut odio euismod &
elementum. Vestibulum tempus.'
lineLenght = 50
beginningIndent = 10
pos1 = 1
n = 0
DO
pos2 = INDEX(text(pos1:), " ")
IF (pos2 == 0) THEN
n = n + 1
words(n) = text(pos1:)
EXIT
END IF
n = n + 1
words(n) = text(pos1:pos1+pos2-2)
pos1 = pos2+pos1
END DO
print*, LEN(words)
i = 0
DO
IF( i >= LEN(words) ) THEN
EXIT
END IF
characterCount = 0
wordsForLinePointer = 0
DO
IF( wordsForLinePointer >= LEN(wordsForLine) ) THEN
EXIT
END IF
wordsForLine(wordsForLinePointer) = ''
wordsForLinePointer = wordsForLinePointer + 1
END DO
wordsForLinePointer = 0
substractForInterval = 1
IF( i == (LEN(wordsForLine) -1) ) THEN
substractForInterval = 0
END IF
DO
IF( (i >= LEN(words)) .AND. (characterCount > lineLenght - LEN(words(i)) - substractForInterval) ) THEN
EXIT
END IF
IF( i == 0 ) THEN
beginningIndentWord = ''
k=0
DO
IF( k >= beginningIndent ) THEN
EXIT
END IF
beginningIndentWord = beginningIndentWord//' '
k = k + 1
END DO
wordsForLine(wordsForLinePointer) = beginningIndentWord
wordsForLinePointer = wordsForLinePointer + 1
characterCount = characterCount + beginningIndent + 1
END IF
wordsForLine(wordsForLinePointer) = words(i)
wordsForLinePointer = wordsForLinePointer + 1
characterCount = characterCount + LEN(words(i)) + 1
i = i + 1
END DO
characterCount = characterCount - 1
numberOfSpacesToAddAfterEachWord = (lineLenght - characterCount) / (LEN(wordsForLine) - 1)
numberOfLeftOverSpaces = MOD((lineLenght - characterCount),(LEN(wordsForLine) - 1))
numberOfLeftOverSpacesAdded = 0
j = 0
DO
IF( j >= (wordsForLinePointer + 1) ) THEN
EXIT
END IF
result = result//wordsForLine(j)
IF( j < wordsForLinePointer ) THEN
result = result//' '
END IF
IF( i /= LEN(words) .AND. j < wordsForLinePointer ) THEN
p = 0
DO
IF( p >= numberOfSpacesToAddAfterEachWord ) THEN
EXIT
END IF
result = result//' '
p = p + 1
END DO
IF( numberOfLeftOverSpacesAdded < numberOfLeftOverSpaces ) THEN
result = result//' '
numberOfLeftOverSpacesAdded = numberOfLeftOverSpacesAdded + 1
END IF
END IF
j = j + 1
END DO
result = result//'\n'
END DO
print*, result
end program xfunc