我正在做所有的K& R练习,我终于能够打印水平直方图了。它看起来也很糟糕,但我会让你判断它。我无法在其输入中垂直打印单词长度的直方图。
如何修改我的程序来执行此操作?
问题:编写程序以打印长度的直方图(如果是单词) 在其输入中。用条形图很容易绘制直方图 水平;垂直方向更具挑战性。
histogram.c
#include <stdio.h>
#define MAX 10
#define IN 1
#define OUT 0
int main()
{
int c, len, state;
int nlength[MAX];
c = len = 0;
state = 1;
for(int i = 0; i < 10; ++i)
nlength[i] = 0;
while ((c = getchar()) != EOF) {
++len;
if (c == ' ' || c == '\n' || c == '\t') {
--len;
state = OUT;
}
if(state == OUT) {
if(len != 0 && len <= MAX)
++nlength[len];
len = 0;
state = IN;
}
}
for (int i = 0; i <= MAX; ++i) {
printf("%d ", i);
for (int a = 0; a < nlength[i]; ++a)
printf("*");
printf("\n");
}
return 0;
}
OUTPUT:
./histogram < histogram.c
0
1 *************************************
2 *************************
3 **************
4 ************
5 *****
6 ******
7 ****
8 **
9 *
10 ***
答案 0 :(得分:3)
首先,你需要知道直方图的高度,它是最大值。然后打印每一行,然后根据值决定放置*
或。
int h_max = 0;
for (int a = 0; a < MAX; a++) {
if (h_max <= nlength[a]) h_max = nlength[a];
}
for (int i = h_max; i >= 0; i--) {
for (int a = 0; a < MAX; ++a) {
if (nlength[a] > i) {
printf("*");
} else {
printf(" ");
}
}
printf("\n");
}
另一种解决方案是水平打印到数组中并按照您想要的方向打印数组。
答案 1 :(得分:1)
让行数等于最高bin值(或bin值的一些其他合适函数)。一次打印一行直方图,对于每一列,根据与该列对应的bin值,决定是否在此行的此列中打印*
。
答案 2 :(得分:1)
//vertical HISTOGRAM2
#include<stdio.h>
#include<stdlib.h>
int main()
{
int c,i,j,arr[10],height=0;
system("clear");
for(i=0 ; i<10 ; i++)
arr[i]=0;
while( ( c=getchar() ) != EOF)
{
if(c >= '0' || c <='9')
++arr[c-'0'];
if( arr[c-'0'] > height )
{
height = arr[c-'0'];
}
}
printf("\n");
for(j=height ; j>0 ; j--) // row
{
printf("%2d|",j);
for ( i=0 ; i<=9 ; i++) // column
{
if( j == arr[i] )
{
printf(" *|");
arr[i]--;
}
else
printf(" |");
}
printf("\n");
}
printf(" |");
for ( i=0 ; i<=9 ; i++)
printf(" %d|",i);
printf("\n ------------DIGITS-------------");
printf("\n");
return(0);
}
答案 3 :(得分:0)
#include <stdio.h>
#define MAX 10
#define IN 1
#define OUT 0
int main(void){
int c, len, state;
int nlength[MAX];
c = len = 0;
state = IN;
for(int i = 0; i < MAX; ++i)
nlength[i] = 0;
while ((c = getchar()) != EOF) {
++len;
if (c == ' ' || c == '\n' || c == '\t') {
--len;
state = OUT;
}
if(state == OUT) {
if(len != 0 && len <= MAX)
++nlength[len-1];
len = 0;
state = IN;
}
}
int max = 0;
//horizontal
for (int i = 0; i < MAX; ++i) {
if(max < nlength[i]) max = nlength[i];
printf("%2d ", i+1);
for (int a = 0; a < nlength[i]; ++a)
printf("*");
printf("\n");
}
printf("\n");
//vertical
for (int i = max; i > 0; --i){
for (int j = 0; j < MAX; ++j)
if(nlength[j]>=i)
printf("%c ", '*');
else
printf("%c ", ' ');
printf("\n");
}
for(int i=1;i<=MAX;++i)
printf("%-2d", i);
return 0;
}
答案 4 :(得分:-1)
#include <stdio.h>
#define MAX_WORDS 10
//Prints a histogram of the lengths of words in its input
int main()
{
int c, numberOfCharacters, numberOfWords, wordLengthsIndex, largestWordLength;
c = numberOfCharacters = numberOfWords = wordLengthsIndex = largestWordLength = 0;
int wordLengths[MAX_WORDS];
//Read the words
while ( ((c = getchar()) != EOF) && numberOfWords < MAX_WORDS)
{
if (c == ' ' || c == '\n' || c == '\t') //no longer in a word
{
if (numberOfCharacters > largestWordLength)
{
largestWordLength = numberOfCharacters;
}
wordLengths[wordLengthsIndex] = numberOfCharacters;
numberOfCharacters = 0; //reset for the next word
numberOfWords++;
wordLengthsIndex++;
}
else //treat anything else as a character in a word
{
numberOfCharacters++;
}
}
//Print the histogram horizontally
for (int numberOfWordsIndex = 0; numberOfWordsIndex < numberOfWords; ++numberOfWordsIndex)
{
for (int numberOfCharactersIndex = 0; numberOfCharactersIndex < wordLengths[numberOfWordsIndex]; ++numberOfCharactersIndex)
{
putchar('*');
}
putchar('\n');
putchar('\n');
}
//Print the histogram vertically
for (int character = largestWordLength; character > 0; --character) //cycle each row
{
for (int word = 0; word < numberOfWords; ++word) //cycle each column
{
if (wordLengths[word] >= character) //this particular word length falls within the current row
{
putchar('*');
putchar('\t');
}
else
{
putchar(' ');
putchar('\t');
}
}
putchar ('\n');
}
return 0;
}