Google Code Jam 2013 – Qualification Round Problem 3


Little John likes palindromes, and thinks them to be fair (which is a fancy word for nice). A palindrome is just an integer that reads the same backwards and forwards – so 6, 11 and 121 are all palindromes, while 10, 12, 223 and 2244 are not (even though 010=10, we don’t consider leading zeroes when determining whether a number is a palindrome).

He recently became interested in squares as well, and formed the definition of a fair and square number – it is a number that is a palindrome and the square of a palindrome at the same time. For instance, 1, 9 and 121 are fair and square (being palindromes and squares, respectively, of 1, 3 and 11), while 16, 22 and 676 are not fair and square: 16 is not a palindrome, 22 is not a square, and while 676 is a palindrome and a square number, it is the square of 26, which is not a palindrome.

Now he wants to search for bigger fair and square numbers. Your task is, given an interval Little John is searching through, to tell him how many fair and square numbers are there in the interval, so he knows when he has found them all.

Solving this problem

Usually, Google Code Jam problems have 1 Small input and 1 Large input. This problem has 1 Small input and 2 Large inputs. Once you have solved the Small input, you will be able to download any of the two Large inputs. As usual, you will be able to retry the Small input (with a time penalty), while you will get only one chance at each of the Large inputs.

Input

The first line of the input gives the number of test cases, T. T lines follow. Each line contains two integers, A and B – the endpoints of the interval Little John is looking at.

Output

For each test case, output one line containing “Case #x: y”, where x is the case number (starting from 1) and y is the number of fair and square numbers greater than or equal to A and smaller than or equal to B.

Limits

Small dataset

1 ≤ T ≤ 100.
1 ≤ A ≤ B ≤ 1000.

First large dataset

1 ≤ T ≤ 10000.
1 ≤ A ≤ B ≤ 1014.

Second large dataset

1 ≤ T ≤ 1000.
1 ≤ A ≤ B ≤ 10100.

Sample Input

3
1 4
10 120
100 1000

Sample Output

Case #1: 2
Case #2: 0
Case #3: 2

My Solution

My brute force approach only solved the small input. After it failed the large ones I realized that a much better approach would be to pre-compute all perfect squares and pre-calculate the number of perfect squares in each range of numbers. Luckily I was already classified.


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

int isPali(long long int n){
  int i;
  int len;
  char number[100];
  sprintf(number,"%lld",n);

  len = strlen(number);
  if(len==1)
    return 1;
  if(len==2){
  
  if(number[0]==number[1])
      return 1;
    else
      return 0;
  }
  else{
    for (i=0;i<len/2;i++){
      if(number[i]!=number[len-1-i])
        return 0;
    }
    return 1;
  }
}

long long int isSquare(long long int n){
  long long int rightBits = n & 0xF;
  long long int square;
  if(rightBits==0||rightBits==1||rightBits==4||rightBits==9){
    square = (long long int)sqrt((double)n);
    if(square*square==n)
      return square;
    else 
      return 0;
  }
  else
    return 0;
}

int processNumbers(long long int min, long long int max){
  long long int i;
  int count = 0;
  long long int sq;

  for(i=min;i<=max;i++){    
    printf("processing %lldn",i);
    sq = isSquare(i);
    if(sq){
      if(isPali(i)&&isPali(sq)){
        count++;
      }
    }
  }

  return count;
}

int main(){
  int k,t;
  long long int a,b;
  int result;
  
  scanf("%d",&t);
  
  for(k=0;k<t;k++){
    scanf("%lld %lld",&a,&b);

    result = processNumbers(a,b);    

    if(k>0)
      printf("n");
    printf("Case #%d: %d",k+1,result);
  }

  return 0;
}

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