plssssss help !!....using array.. turbo c.. create a program that will acce

plssssss help !!....using array.. turbo c..

create a program that will accept number of words to be
consored.

.a word must not exceed 10 characters long
.the text to be entered will be no longer than 200 characters
.there will be no 10 words

example:

enter number of words to be censor: 5

enter words to censor:

windows
office
microsoft
bill
gates

enter text to censor:

bill gates founded microsoft and makes office and windows

sample output:

<consored> <censored> founded <censored> and makes
<censored> and <censored>

Question Posted / hmmm

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plssssss help !!....using array.. turbo c.. create a program that will accept number of words t..

Answer / swapnil chhajer

#include<stdio.h>
#include<stdlib.h>
#include<string.h>

int main()
{
int nocw,i,j,flag=0;
char
cenWords[20][12],text[205],temp[12],finalText[450]={'\0'};

printf("Enter number of words to be censor : ");
scanf("%d",&nocw);

printf("\nEnter words to be censor : ");
for(i=0;i<nocw;i++)
scanf("%s",cenWords[i]);

fflush(stdin);
printf("\nEnter text to censor : ");
gets(text);

for(i=0;i<strlen(text);i++)
{
j=0;
flag=0;
while(!(text[i]==' '||text[i]=='\t'||text[i]=='\n'))
{
temp[j++]=text[i++];
}
temp[j]='\0';

for(j=0;j<nocw;j++)
{
if(strcmp(temp,cenWords[j])==0)
{
strcat(finalText,"<censored> ");
flag=1;
break;
}
}

if(flag==0)
{
strcat(finalText,temp);
strcat(finalText," ");
}
}

printf("\n\n :: FINAL TEXT :: \n\n");
puts(finalText);
getchar();
return 0;
}

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6 Yes

4 No

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