Scanf() function isn't taking string input - c

Question

After taking test case inputs it does not waits for string input. I have tried fgets(); But, fgets() requires a newline from user which i think will not be accepted by online judge. So, whats the problem here ? Here is my code.
#include<stdio.h>
int main ()
{
int t, j;
int cnt = 0;
scanf("%d", &t);
while(t--)
{
cnt++;
int i, count = 0;
char s[101];
scanf("%[^\n]s", s);
for(i=0;s[i] != '\0'; i++)
{
if(s[i] == 'a' || s[i] == 'd' || s[i] == 'g' || s[i] == 'j' || s[i] == 'm' || s[i] == 'p' || s[i] == 't' || s[i] == 'w' || s[i] == ' ')
{
count += 1;
continue;
}
else if(s[i] == 'b' || s[i] == 'e' || s[i] == 'h' || s[i] == 'k' || s[i] == 'n' || s[i] == 'q' || s[i] == 'u' || s[i] == 'x')
{
count += 2;
continue;
}
else if(s[i] == 'c' || s[i] == 'f' || s[i] == 'i' || s[i] == 'l' || s[i] == 'o' || s[i] == 'r' || s[i] == 'v' || s[i] == 'y')
{
count += 3;
continue;
}
else if(s[i] == 's' || s[i] == 'z')
{
count += 4;
continue;
}
}
printf("Case #%d: %d\n", cnt, count);
}
return 0;
}

Answer 1

Change
scanf("%[^\n]s", s);
to
scanf(" %100[^\n]", s);
The space at the beginning makes it skip over any whitespace before the word. This is needed to allow it to read past the newline at the end of each line before scanning the next line.
100 ensures that it doesn't try to write past the end of the string.
You shouldn't have s at the end, because that will try to match a literal s in the input. [^\n] is not a modifier of the %s format operator, it's a completely different format operator.

Answer 2

It would help immensely if you had included a link to the specific question. Then we could advise if you had answered the question correctly.
For instance, what about capital letters, punctuation, etc.?
For instance, does the question guarantee that each input string will <= 100 characters, or is there a criteria that no more than the first 100 characters be used?
We cannot determine if the posted code is answering the question if we do not know what the question is.
Most I/O is very time/CPU cycles expensive, especially scanf() and printf()
Most online questions have the execution timed and exceeding the allowed time limit will cause the proposed answer (that you provide) to fail.
Appropriate 'fast' I/O functions are easily found by YOU by looking at the answers to prior questions. Here are a couple of fully functional examples to get you started:
#include <stdio.h>
void fastRead( size_t *a );
void fastWrite( size_t a );
inline void fastRead(size_t *a)
{
int c=0;
// note: 32 is space character
while (c<33) c=getchar_unlocked();
// initialize result value
*a=0;
// punctuation parens, etc are show stoppers
while (c>47 && c<58)
{
*a = (*a)*10 + (size_t)(c-48);
c=getchar_unlocked();
}
//printf( "%s, value: %lu\n", __func__, *a );
} // end function: fastRead
inline void fastWrite(size_t a)
{
char snum[20];
//printf( "%s, %lu\n", __func__, a );
int i=0;
do
{
// 48 is numeric character 0
snum[i++] = (char)((a%10)+(size_t)48);
a=a/10;
}while(a>0);
i=i-1; // correction for overincrement from prior 'while' loop
while(i>=0)
{
putchar_unlocked(snum[i--]);
}
putchar_unlocked('\n');
} // end function: fastWrite
Most often (you may have to experiment a bit) the online code judging wants a newline output to stdout after each test case, including the last test case. This is usually accomplished by:
putchar_unlocked('\n');
When a value will never be less than 0, it is (usually) best to use size_t rather than int,
It is good programming practice to use variable names that indicate content or usage (or better, both).
code is MUCH more readable and understandable if
it is consistently indented
code blocks are separated by a single blank line
meaningful variable names are used
all 'distractor' code is eliminated
Putting all the above together results in:
#include <stdio.h> // putchar_unlocked(), getchar_unlocked()
void fastRead( size_t *a );
void fastWrite( size_t a );
inline void fastRead(size_t *a)
{
int c=0;
// note: 32 is space character
while (c<33) c=getchar_unlocked();
// initialize result value
*a=0;
// punctuation parens, etc are show stoppers
while (c>47 && c<58)
{
*a = (*a)*10 + (size_t)(c-48);
c=getchar_unlocked();
}
//printf( "%s, value: %lu\n", __func__, *a );
} // end function: fastRead
inline void fastWrite(size_t a)
{
char snum[20];
//printf( "%s, %lu\n", __func__, a );
int i=0;
do
{
// 48 is numeric character 0
snum[i++] = (char)((a%10)+(size_t)48);
a=a/10;
}while(a>0);
i=i-1; // correction for overincrement from prior 'while' loop
while(i>=0)
{
putchar_unlocked(snum[i--]);
}
putchar_unlocked('\n');
} // end function: fastWrite
int main ( void )
{
size_t numTestCases;
fastRead( &numTestCases );
for( size_t testCase =1; testCase <= numTestCases; testCase++)
{
size_t count = 0;
int ch;
while( (ch = getchar_unlocked()) != '\n' )
{
if( ch == 'a'
|| ch == 'd'
|| ch == 'g'
|| ch == 'j'
|| ch == 'm'
|| ch == 'p'
|| ch == 't'
|| ch == 'w'
|| ch == ' ')
{
count += 1;
}
else if( ch == 'b'
|| ch == 'e'
|| ch == 'h'
|| ch == 'k'
|| ch == 'n'
|| ch == 'q'
|| ch == 'u'
|| ch == 'x')
{
count += 2;
}
else if( ch == 'c'
|| ch == 'f'
|| ch == 'i'
|| ch == 'l'
|| ch == 'o'
|| ch == 'r'
|| ch == 'v'
|| ch == 'y')
{
count += 3;
}
else if( ch == 's'
|| ch == 'z')
{
count += 4;
}
}
printf("Case #%lu: %lu\n", testCase, count);
putchar_unlocked( '\n' );
}
return 0;
}
The call to printf() could also be greatly sped up by using a series of calls to fastWrite() and putchar_unlocked()
Also, unless the question specifically asks for the superfluous characters being used in the call to printf() the code probably should say:
fastWrite( testCase );
putchar_unlocked( ' ' );
fastWrite( count );
putchar_unlocked( '\n' );
it is also highly desirable to list the parameters of the question at the top of your code. This both helps you and anyone else reading your code.
Here is a trivial example of such documentation:
/*
* criteria 1
* Using two characters: . (dot) and * (asterisk)
* print a frame-like pattern
*
* criteria 2
* You are given t - the number of test cases
*
* criteria 3
* for each of the test cases input two positive integers:
* l - the number of lines and
* c - the number of columns of a frame.
*
* criteria 4
* Use one line break in between successive patterns
*/

Answer 3

Given the latest information you have provided, here is the answer that I would suggest.
If it still does not pass the judge, at least you have a very solid base from which to experiment to determine the correct answer.
/*
* Cell phones have become an essential part of modern life.
* In addition to making voice calls,
* cell phones can be used to send text messages,
* which are known as SMS for short.
* Unlike computer keyboards, most cell phones have limited number of keys.
* To accommodate all alphabets, letters are compacted into single key.
* Therefore, to type certain characters,
* a key must be repeatedly pressed until that character is shown on the display
panel.
*
* In this problem we are interested in finding out the number of times
* keys on a cell phone must be pressed to type a particular message.
* In this problem we will assume that the key pad of our cell phone is arranged as follows.
* ---------------------
* |1 | abc | def |
* ---------------------
* | ghi | jkl | mno |
* ---------------------
* | pqrs | tuv | wxyz |
* ---------------------
* | | <SP>| |
* ---------------------
* In the above grid each cell represents one key.
* Here <SP> means a space.
* In order to type the letter `a', we must press that key once,
* however to type `b' the same key must be pressed twice
* and for `c' three times.
* In the same manner, one key press for `d',
* two for `e'
* and three for `f'.
* This is also applicable for the remaining keys and letters.
* Note that it takes a single press to type a space.
*
* Input
* The first line of input will be a positive integer T
* where T denotes the number of test cases.
* T lines will then follow each containing only spaces and lower case letters.
* Each line will contain at least 1 and at most 100 characters.
*
* Output
* For every case of input there will be one line of output.
* It will first contain the case number
* followed by the number of key presses required to type the message of that case.
* Look at the sample output for exact formatting.
*
* Sample Input
* 2
* welcome to ulab
* good luck and have fun
*
* Sample Output
*
* Case #1: 29
* Case #2: 41
*/
#include <stdio.h> // putchar_unlocked(), getchar_unlocked(), puts(), sprintf()
#include <string.h> // strcpy(), strcat()
#define MAX_OUTPUT_LEN 50
void fastRead( size_t *a );
void fastWrite( size_t a );
inline void fastRead(size_t *a)
{
int c=0;
// note: 32 is space character
while (c<33) c=getchar_unlocked();
// initialize result value
*a=0;
// punctuation parens, etc are show stoppers
while (c>47 && c<58)
{
*a = (*a)*10 + (size_t)(c-48);
c=getchar_unlocked();
}
//printf( "%s, value: %lu\n", __func__, *a );
} // end function: fastRead
inline void fastWrite(size_t a)
{
char snum[20];
//printf( "%s, %lu\n", __func__, a );
int i=0;
do
{
// 48 is numeric character 0
snum[i++] = (char)((a%10)+(size_t)48);
a=a/10;
}while(a>0);
i=i-1; // correction for overincrement from prior 'while' loop
while(i>=0)
{
putchar_unlocked(snum[i--]);
}
putchar_unlocked('\n');
} // end function: fastWrite
const size_t keyPressCounts[] =
{
1, 2, 3, // a b c
1, 2, 3, // d e f
1, 2, 3, // g h i
1, 2, 3, // j k l
1, 2, 3, // m n o
1, 2, 3, 4,// p q r s
1, 2, 3, // t u v
1, 2, 3, 4 // w x y z
};
int main ( void )
{
size_t numTestCases;
fastRead( &numTestCases );
for( size_t testCase =1; testCase <= numTestCases; testCase++)
{
size_t keypresses = 0;
int ch;
while( (ch = getchar_unlocked()) != '\n' )
{
if( ' ' == ch )
keypresses += 1;
else
keypresses += keyPressCounts[ ch - 'a' ];
} // end while()
char outputLine[ MAX_OUTPUT_LEN ];
strcpy( outputLine, "Case #" );
sprintf( &outputLine[ strlen(outputLine) ], "%lu", testCase );
strcat( outputLine, ": " );
sprintf( &outputLine[ strlen(outputLine) ], "%lu", keypresses );
puts( outputLine );
} // end for()
return 0;
} // end function: main