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 smallest digit is 1
So, 3*11+7*1=40, so 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
Answers
Answer:
cannnnot
Explanation:
understand okkkkkkkk
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Answer:
The program is done in c language
Explanation:
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 denotes the number of bit pairs.
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 number of pairs possible is 3.
#include<iostream>
using namespace std;
int bit_score(int n)
{
int a, b, c, largest, smallest;
int score;
a = n%10; n/=10;
b = n%10; n/=10;
c = n%10; n/=10;
largest = (a>b)?a:b;
largest = (c>largest)?c:largest;
smallest = (a<b)?a:b;
smallest = (c<smallest)?c:smallest;
score = largest*11 + smallest*7;
score = score % 100;
return score;
}
int findPairs (int score_array[], int N)
{
int sig_dig[10], i, pairs = 0, msb;
for(i=0; i<10; i++)
{
sig_dig[i] = 0;
}
for(i=0; i<N; i=i+2)
{
msb = score_array[i] / 10;
for(int j =i+2; j<N; j=j+2)
{
if(msb == score_array[j]/10)
{
if(sig_dig[msb] < 2)
{
sig_dig[msb]++;
}
}
}
}
for(i=1; i<N; i=i+2)
{
msb = score_array[i] / 10;
for(int j =i+2; j<N; j=j+2)
{
if(msb == score_array[j]/10)
{
if(sig_dig[msb] < 2)
{
sig_dig[msb]++;
}
}
}
}
for(i=0; i<10; i++)
{
pairs = pairs + sig_dig[i];
}
return pairs;
}
int main()
{
int N, i;
int ip_array[501];
int score_array[501];
int pairs;
cin>>N;
for(i=0; i<N; i++)
{
cin>>ip_array[i];
}
for(i=0; i<N; i++)
{
score_array[i] = bit_score(ip_array[i]);
}
pairs = findPairs(score_array, N);
cout<<pairs;
return 0;
}