Solução: Problema da escrita no celular (Cátia Souza)
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Revision as of 12:27, 1 April 2009 by 200.17.147.2 (Talk) (New page: <code lang="c"> #include <stdio.h> #include <stdlib.h> int EhNumero(char num){ char Numeros[] = {'2','3','4','5','6','7','8','9'}; int i; for(i=0;i<=7;i++){ if(num==...)
#include <stdio.h> #include <stdlib.h> int EhNumero(char num){ char Numeros[] = {'2','3','4','5','6','7','8','9'}; int i; for(i=0;i<=7;i++){ if(num==Numeros[i]){ return num; } } return -1; } int main(int argc, char** argv) { char Teclas[]={'2','a','b','c','3','d','e','f','4','g','h','i','5','j','k','l','6','m','n','o','7','p','q','r','s','8','t','u','v','9','w','x','y','z'}; char numero; char palavra[30]; int i,count=0, casou,j; gets(palavra); for(i=0; i< strlen(palavra);i++){ casou=0; for(j=0;j<=34 && (casou!=1);j++){ if(EhNumero(Teclas[j])!=-1){ numero=Teclas[j]; count=0; } if(palavra[i] != Teclas[j]){ count++; } else{ casou=1; printf("#%c=%d\n",numero,count); } } } return (EXIT_SUCCESS); }