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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
4

2 回答 2

0
      CHARACTER*1000 TEXT,TEXTOUT
      TEXTOUT=""
      L=0
      TEXT=" your Latin text, split over n lines using the continuation charactaer &...."
      I=1
   99 TEXTOUT(L+1:L+1)=TEXT(I,I)
      L=L+1
      IF(TEXT(I:I).NE." ")THEN
      I=I+1
      IF(I.LE.LEN_TRIM(TEXT))GO TO 99
      ELSE     
      J=I
      ISPACE=0 
   98 IF(TEXT(J:J).EQ." ")THEN
      J=J+1
      IF(J.LE.LEN_TRIM(TEXT))GO TO 98
      ELSE
      I=J
      GO TO 99
      ENDIF
      ENDIF   
于 2013-06-18T04:01:34.870 回答
0

Fortran 现在有可变长度的字符串。有一个模块添加了这些,但不是很受欢迎。它们已作为可分配缩放器添加到 Fortran 2003。例如,参见Fortran: Syntax error in CHARACTER declaration

于 2013-05-02T12:27:55.887 回答