Question

In: Computer Science

Please make my Code working and pass the test but do NOT change anything in main...

Please make my Code working and pass the test but do NOT change anything in main function, thank you.

#include <iostream>

using namespace std;

void sort(int *A, int n){

  

for(int passes = 0;passes < 2;passes++)

{

// shift can have only two values either 0 or 16, used for shifting purpose

int shift = passes * 16;

int N = 1<<(16 + 1);

  

// Temporary array for storing frequency of upper or lower 16 bits

int temp[N];

  

// initialze all value to zero at first

for(int i = 0;i < N;i++)

temp[i] = 0;

// storing the frequecy of either lower or upper half the every number

for(int i = 0;i < n;i++){

temp[(A[i] >> shift)&0xFFFF]++;

}

// making cummulative sum of the frequecies

for(int i = 1;i < N;i++)

temp[i] += temp[i-1];

// A temporary output array for storing ns

int output[n];

  

// storing the number in their requied position

for(int i = n - 1;i >= 0;i--)

{

output[ temp[(A[i] >> shift)&(0xFFFF)] - 1] = A[i];

temp[ (A[i] >> shift)&(0xFFFF) ]--;

}

  

// copy the values of temporary output array to original array

for(int i = 0;i < n;i++){

A[i] = output[i];

}

}

}

void sort( int *A, int n);

int main()

{

int i, offset, j;

int B[10000000];

time_t t;

srand( (unsigned) time( &t ));

offset = rand()%10000000;

for( i = 0; i< 10000000; i++ )

{

B[i] = ((91*i)%10000000) + offset;

}

printf("Prepared array of 10 million integers; calling sort\n");

sort( B, 10000000);

printf("finished sort, now check result\n");

for( i=0, j=0; i < 10000000; i++ )

if( B[i] != i+ offset ) j++;

if( j == 0 )

printf("Passed Test\n");

else

printf("Failed Test. %d numbers wrong.\n", j );

}

Solutions

Expert Solution

Solution- This code works successfully in Gcc 6.3, C++14 environment. Maybe, there is issue in the compiler itself.

Screenshot of code and IO -

In the following screenshot it is clear that the code gives the output that shows test cases passed.

Code

#include <iostream>

using namespace std;

void sort(int *A, int n){

  

for(int passes = 0;passes < 2;passes++)

{

// shift can have only two values either 0 or 16, used for shifting purpose

int shift = passes * 16;

int N = 1<<(16 + 1);

  

// Temporary array for storing frequency of upper or lower 16 bits

int temp[N];

  

// initialze all value to zero at first

for(int i = 0;i < N;i++)

temp[i] = 0;

// storing the frequecy of either lower or upper half the every number

for(int i = 0;i < n;i++){

temp[(A[i] >> shift)&0xFFFF]++;

}

// making cummulative sum of the frequecies

for(int i = 1;i < N;i++)

temp[i] += temp[i-1];

// A temporary output array for storing ns

int output[n];

  

// storing the number in their requied position

for(int i = n - 1;i >= 0;i--)

{

output[ temp[(A[i] >> shift)&(0xFFFF)] - 1] = A[i];

temp[ (A[i] >> shift)&(0xFFFF) ]--;

}

  

// copy the values of temporary output array to original array

for(int i = 0;i < n;i++){

A[i] = output[i];

}

}

}

void sort( int *A, int n);

int main()

{

int i, offset, j;

int B[10000000];

time_t t;

srand( (unsigned) time( &t ));

offset = rand()%10000000;

for( i = 0; i< 10000000; i++ )

{

B[i] = ((91*i)%10000000) + offset;

}

printf("Prepared array of 10 million integers; calling sort\n");

sort( B, 10000000);

printf("finished sort, now check result\n");

for( i=0, j=0; i < 10000000; i++ )

if( B[i] != i+ offset ) j++;

if( j == 0 )

printf("Passed Test\n");

else

printf("Failed Test. %d numbers wrong.\n", j );

}

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