Answer Posted / abdur rab

#include <stdio.h>

char* str_str ( char* cp_str, char* cp_pattern )
{
char* cp_temp = NULL;
int n_pattern_length = 0;

n_pattern_length = strlen ( cp_pattern );

while ( cp_str && *cp_str ) {
if ( !strncmp ( cp_str, cp_pattern, (
n_pattern_length ) ) ) {
cp_temp = cp_str;
break;
}
else cp_str++;
}

return ( cp_temp );

}

int main ( int argc, char* argv [ ] )
{
char array [] = {"Hello World"};
char* cp_temp = NULL;

cp_temp = str_str ( array, "lo " );
if ( NULL != cp_temp ) {
printf ("\n%s", cp_temp);
} else printf ("\nReturned null");

return ( 0 );

}

In cryptography, you could often break the algorithm if you know what was the original (plain) text that was encoded into the current ciphertext. This is called the plain text attack. In this simple problem, we illustrate the plain text attack on a simple substitution cipher encryption, where you know each letter has been substituted with a different letter from the alphabet but you don’t know what that letter is. You are given the cipherText as the input string to the function getwordSets(). You know that a plain text "AMMUNITION" occurs somewhere in this cipher text. Now, you have to find out which sets of characters corresponds to the encrypted form of the "AMMUNITION". You can assume that the encryption follows simple substitution only. [Hint: You could use the pattern in the "AMMUNITION" like MM occurring twice together to identify this]

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