Advertisement
Not a member of Pastebin yet?
Sign Up,
it unlocks many cool features!
- #include <stdlib.h>
- #include <stdio.h>
- #include <string.h>
- #define WORDS 10
- #define COL 50
- #define ASCII_A 97
- #define ASCII_Z 122
- #define NUM_LETTERS 26
- //This code is taking the words that the user entered and checks if the sentence is pangram.
- int main(void)
- {
- int notYet = 1; // It's showing if it's already a pangram.
- int englet = 0;
- int engLetter = 0;
- int word = 0;
- int letter = 0;
- int i = 0;
- int j = 0;
- char englishLetters[NUM_LETTERS] = {0};
- char pangram[WORDS][COL] = {0};
- printf("Enter up to 10 words, try to make a pangram:\n");
- for(i = 0; (i < WORDS) && (notYet > 0); i++)
- {
- for(j = 0; j < NUM_LETTERS; j++)
- {
- englishLetters[j] = 1;
- }
- fgets(pangram[i], WORDS, stdin);
- if(strlen(pangram[i]) < COL)
- {
- pangram[i][strlen(pangram[i])-1] = 0;
- }
- for(englet = ASCII_A; englet <= ASCII_Z; englet++)
- {
- for(word = 0; word < WORDS; word++)
- {
- for(letter = 0; letter < strlen(pangram[word]); letter++)
- {
- if((pangram[word][letter] == englet) && (englishLetters[engLetter] != 0))
- {
- englishLetters[engLetter] = 0;
- }
- }
- }
- }
- notYet = 0;
- for(englet = 0; (englet > NUM_LETTERS) && (notYet == 0); englet++)
- {
- if(englishLetters[englet] != 0)
- {
- notYet++;
- printf("\n");
- }
- }
- engLetter++;
- }
- if(notYet == 0)
- {
- printf("It's a pangram?\nYes!");
- }
- else
- {
- printf("It's a pangram?\nNo");
- }
- return(0);
- }
Advertisement
Add Comment
Please, Sign In to add comment
Advertisement