Day: September 10, 2015

Flipping bits

Problem Statement

You will be given a list of 32 bits unsigned integers. You are required to output the list of the unsigned integers you get by flipping bits in its binary representation (i.e. unset bits must be set, and set bits must be unset).

Input Format

The first line of the input contains the list size T, which is followed by T lines, each line having an integer from the list.

Constraints

1≤T≤100
0≤integer<232

Output Format

Output one line per element from the list with the requested result.

Sample Input

3 
2147483647 
1 
0

Sample Output

2147483648 
4294967294 
4294967295

Explanation

Take 1 for example, as unsigned 32-bits is 00000000000000000000000000000001 and doing the flipping we get 11111111111111111111111111111110 which in turn is 4294967294.

Solution

  
#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;


int main() {
    /* Enter your code here. Read input from STDIN. Print output to STDOUT */
    int t,n;
    unsigned int a;
    cin>>t;
    while(t--)
    {
        cin>>a;
        cout<<~a<<endl;      
    }
    return 0;
    
}

Funny String

Problem Statement

Suppose you have a string S which has length N and is indexed from 0 to N−1. String R is the reverse of the string S. The string S is funny if the condition |Si−Si−1|=|Ri−Ri−1| is true for every i from 1 to N−1.

(Note: Given a string str, stri denotes the ascii value of the ith character (0-indexed) of str. |x| denotes the absolute value of an integer x)

Input Format

First line of the input will contain an integer T. T testcases follow. Each of the next T lines contains one string S.

Constraints

  • 1<=T<=10
  • 2<=length of S<=10000

Output Format

For each string, print Funny or Not Funny in separate lines.

Sample Input

2
acxz
bcxz

Sample Output

Funny
Not Funny

Explanation

Consider the 1st testcase acxz

here

|c-a| = |x-z| = 2
|x-c| = |c-x| = 21
|z-x| = |a-c| = 2

Hence Funny.

Consider the 2nd testcase bcxz

here

|c-b| != |x-z|

Hence Not Funny.

Solution

  
#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;


int main() {
    /* Enter your code here. Read input from STDIN. Print output to STDOUT */ 
    int t,count;
    string s,r;
    cin>>t;
    while(t--)
    {
        count=0;
        r.clear();
        s.clear();
        cin>>s;
        r=s;
        reverse(r.begin(),r.end());
        for(int i=1;i<s.length();i++)
        {
            if(abs(s[i]-s[i-1])==abs(r[i]-r[i-1]))
            {
                count++;
            }
        }
        if(count==(s.length()-1))
            cout<<"Funny"<<endl;
        else cout<<"Not Funny"<<endl;
    }
    return 0;
}

Pangrams

Problem Statement

Roy wanted to increase his typing speed for programming contests. So, his friend advised him to type the sentence “The quick brown fox jumps over the lazy dog” repeatedly, because it is a pangram. (Pangrams are sentences constructed by using every letter of the alphabet at least once.)

After typing the sentence several times, Roy became bored with it. So he started to look for other pangrams.

Given a sentence s, tell Roy if it is a pangram or not.

Input Format Input consists of a line containing s.

Constraints
Length of s can be at most 103 (1≤|s|≤103) and it may contain spaces, lower case and upper case letters. Lower case and upper case instances of a letter are considered the same.

Output Format Output a line containing pangram if s is a pangram, otherwise output not pangram.

Sample Input #1

We promptly judged antique ivory buckles for the next prize

Sample Output #1

pangram

Sample Input #2

We promptly judged antique ivory buckles for the prize

Sample Output #2

not pangram

Explanation
In the first test case, the answer is pangram because the sentence contains all the letters of the English alphabet.

Solution in Python: 

  
# Enter your code here. Read input from STDIN. Print output to STDOUT
print 'pangram' if set(raw_input().lower().replace(' ', '')) == set('abcdefghijklmnopqrstuvwxyz') else 'not pangram'