Digit Pairs Problem Description Given N three-digit numbers, your task is to find bit score of all N numbers and then print the number of pairs possible based on these calculated bit score. 1. Rule for calculating bit score from three digit number: From the 3-digit number, · extract largest digit and multiply by 11 then · extract smallest digit multiply by 7 then · add both the result for getting bit pairs. Note: - Bit score should be of 2-digits, if above results in a 3-digit bit score, simply ignore most significant digit. Consider following examples: Say, number is 286 Largest digit is 8 and smallest digit is 2 So, 8*11+2*7 =102 so ignore most significant bit , So bit score = 02. Say, Number is 123 Largest digit is 3 and the smallest digit is 1 So, 3*11+7*1=40, so a bit score is 40. 2. Rules for making pairs from above calculated bit scores Condition for making pairs are · Both bit scores should be in either odd position or even position to be eligible to form a pair. · Pairs can be only made if most significant digit are same and at most two pair can be made for a given significant digit. Constraints N<=500 Input Format First line contains an integer N, denoting the count of numbers. Second line contains N 3-digit integers delimited by space Output One integer value denoting the number of bit pairs. Timeout 1 Explanation Example 1 Input 8 234 567 321 345 123 110 767 111 Output 3 Explanation After getting the most and least significant digits of the numbers and applying the formula given in Rule 1 we get the bit scores of the numbers as: 58 12 40 76 40 11 19 18 No. of pair possible are 3: 40 appears twice at odd-indices 3 and 5 respectively. Hence, this is one pair. 12, 11, 18 are at even-indices. Hence, two pairs are possible from these three-bit scores. Hence total pairs possible is 3
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Answer:
Digit Pairs Problem Description Given N three-digit numbers, your task is to find bit score of all N numbers and then print the number of pairs possible based on these calculated bit score. 1. Rule for calculating bit score from three digit number: From the 3-digit number, · extract largest digit and multiply by 11 then · extract smallest digit multiply by 7 then · add both the result for getting bit pairs. Note: - Bit score should be of 2-digits, if above results in a 3-digit bit score, simply ignore most significant digit. Consider following examples: Say, number is 286 Largest digit is 8 and smallest digit is 2 So, 8*11+2*7 =102 so ignore most significant bit , So bit score = 02. Say, Number is 123 Largest digit is 3 and the smallest digit is 1 So, 3*11+7*1=40, so a bit score is 40. 2. Rules for making pairs from above calculated bit scores Condition for making pairs are · Both bit scores should be in either odd position or even position to be eligible to form a pair. · Pairs can be only made if most significant digit are same and at most two pair can be made for a given significant digit. Constraints N<=500 Input Format First line contains an integer N, denoting the count of numbers. Second line contains N 3-digit integers delimited by space Output One integer value denoting the number of bit pairs. Timeout 1 Explanation Example 1 Input 8 234 567 321 345 123 110 767 111 Output 3 Explanation After getting the most and least significant digits of the numbers and applying the formula given in Rule 1 we get the bit scores of the numbers as: 58 12 40 76 40 11 19 18 No. of pair possible are 3: 40 appears twice at odd-indices 3 and 5 respectively. Hence, this is one pair. 12, 11, 18 are at even-indices. Hence, two pairs are possible from these three-bit scores. Hence total pairs possible is 3
Explanation:
Digit Pairs Problem Description Given N three-digit numbers, your task is to find bit score of all N numbers and then print the number of pairs possible based on these calculated bit score. 1. Rule for calculating bit score from three digit number: From the 3-digit number, · extract largest digit and multiply by 11 then · extract smallest digit multiply by 7 then · add both the result for getting bit pairs. Note: - Bit score should be of 2-digits, if above results in a 3-digit bit score, simply ignore most significant digit. Consider following examples: Say, number is 286 Largest digit is 8 and smallest digit is 2 So, 8*11+2*7 =102 so ignore most significant bit , So bit score = 02. Say, Number is 123 Largest digit is 3 and the smallest digit is 1 So, 3*11+7*1=40, so a bit score is 40. 2. Rules for making pairs from above calculated bit scores Condition for making pairs are · Both bit scores should be in either odd position or even position to be eligible to form a pair. · Pairs can be only made if most significant digit are same and at most two pair can be made for a given significant digit. Constraints N<=500 Input Format First line contains an integer N, denoting the count of numbers. Second line contains N 3-digit integers delimited by space Output One integer value denoting the number of bit pairs. Timeout 1 Explanation Example 1 Input 8 234 567 321 345 123 110 767 111 Output 3 Explanation After getting the most and least significant digits of the numbers and applying the formula given in Rule 1 we get the bit scores of the numbers as: 58 12 40 76 40 11 19 18 No. of pair possible are 3: 40 appears twice at odd-indices 3 and 5 respectively. Hence, this is one pair. 12, 11, 18 are at even-indices. Hence, two pairs are possible from these three-bit scores. Hence total pairs possible is 3
Answer:
Problem: Find gcd(m,n), the greatest common divisor of
two nonnegative, not both zero integers m and n
Examples: gcd(60,24) = 12, gcd(60,0) = 60,gcd(0,0) = ?
Euclid’s algorithm is based on repeated application of
equality gcd(m,n) = gcd(n, m mod n) until the second
number becomes 0, which makes the
Problem trivial.
Example: gcd(60,24) = gcd(24,12) = gcd(12,0) = 12
//The algorithm sorts a given array by selection sort
//Input: An array A[0..n-1] of orderable elements
//Output: Array A[0..n-1] sorted in ascending order
#SPJ2