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C program to count numbers of positive and negative numbers

Write a C program using arrays to count the numbers of positive and negative numbers which accepts the inputs as "size of the array" & "elements of the array" and outputs as "number of negative numbers in an array are" & "numbers of positive numbers in an array are" ?
Asked by: Interview Candidate | Asked on: Oct 12th, 2007
Showing Answers 1 - 7 of 7 Answers
baseersd

Answered On : Oct 31st, 2007

View all answers by baseersd

Code
  1. #include
  2. int main()
  3. {
  4.     int p_nArr[20]={0};    
  5.     int nCount_Pos = 0;
  6.     int nCount_Neg = 0;
  7.     int num=0;
  8.     int SIZEOFARRAY= 0;
  9.     int index;
  10.     printf("nEnter the Size of the array");
  11.     scanf("%d",&SIZEOFARRAY);
  12.     printf("nEnter the elements ::n");
  13.     for( index = 0; index < SIZEOFARRAY; index++ )
  14.     {
  15.          scanf("%d",&p_nArr[index]);
  16.          if (p_nArr[index] > 0 )
  17.             nCount_Pos++ ;
  18.          else if (p_nArr[index] < 0 )
  19.              nCount_Neg++;
  20.     }
  21.     if ( nCount_Neg > 0)
  22.     {
  23.        printf("nNumber of negative numbers in the array are :: %dn", nCount_Neg);
  24.        for( index = 0; index < SIZEOFARRAY; index++ )      
  25.        {
  26.           if (p_nArr[index] < 0 )  
  27.          printf("%d ",p_nArr[index]);            
  28.        }
  29.     }
  30.     if ( nCount_Pos > 0 )
  31.     {
  32.            printf("nNumber of Positive numbers in the array are :: %dn",nCount_Pos);
  33.            for( index = 0; index < SIZEOFARRAY; index++ )      
  34.            {
  35.               if (p_nArr[index] > 0 )  
  36.              printf("%d ",p_nArr[index]);            
  37.            }          
  38.     }
  39.  
  40.     getch();
  41.  
  42. }
  43.  
  44. }

//Note : This program can be solved efficiently using linked list.

  
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kaushalgoa

Answered On : Dec 27th, 2007

View all answers by kaushalgoa

Code
  1. void func(int num, int * arr, int *pos, int *neg)
  2. {
  3.     int i = 0;
  4.     *pos = 0;
  5.     *neg = 0;
  6.     for(i = 0; i < num; i++)
  7.     {
  8.         if(*(arr + i) >= 0)
  9.             (*pos)++;
  10.         else
  11.             (*neg)++;
  12.     }
  13.     return;
  14. }
  15.  
  16. void main()
  17. {
  18.    
  19.     int array[] = {1, 2, 3, 4, 5, -4, -6, -8, 0};
  20.     int pos, neg;
  21.     func(9, array, &pos, &neg);
  22.     return;
  23.    
  24. }

  
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jintojos

Answered On : Jun 18th, 2008

View all answers by jintojos

Code
  1.  
  2. #include<stdio.h>
  3. #include<conio.h>
  4.  
  5. void main()
  6.  {
  7.     int *arr,num,i;
  8.     clrscr();
  9.     printf("Enter The Number Of Elements :");
  10.     scanf("%d",&num);
  11.     arr=(int *)malloc(num*2);
  12.     printf("Enter The numbers :");
  13.     for(i=0;i<num;i++) scanf("%d",(arr+i));
  14.     printf("nPositive Numbers Are....n");
  15.     for(i=0;i<num;i++) if(arr[i]>0) printf(" %d",arr[i]);
  16.     printf("nNegative Numbers Are....n");
  17.     for(i=0;i<num;i++) if(arr[i]<0) printf(" %d",arr[i]);
  18.     getch();
  19.  }
  20.  
  21.  

  
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jintojos

Answered On : Jun 18th, 2008

View all answers by jintojos

Code
  1.  
  2. #include<conio.h>
  3. #include<stdio.h>
  4.  
  5. void main()
  6.   {
  7.       int *arr,num,i=0,pos=0,neg=0;
  8.       clrscr();
  9.       printf("Enter The Number Of Elements :");
  10.       scanf("%d",&num);
  11.       arr=(int *)malloc(num*2);
  12.       printf("Enter The Numbers :");
  13.       for(i=0;i<num;i++)
  14.          {
  15.             scanf("%d",arr+i);  
  16.             if(arr[i]>0) pos++;
  17.                 else neg++;
  18.          }
  19.       printf("The Number Of Positive Numbers Are :%d",pos);
  20.       printf("The Number Of Negative Numbers Are :%d",neg);
  21.       printf("Number Of Zeros Are :%d",num-(pos+neg));
  22.       getch();
  23.   }
  24.  
  25.  

  
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kesineni

Answered On : Jul 12th, 2009

View all answers by kesineni

Code
  1.  
  2. #include <stdio.h>
  3.  
  4. main()
  5. {
  6.    int a[] = { 1,-3,-5,-4,-9,2,8};
  7.    int len,i,val;
  8.    int npos=0, nneg=0;
  9.  
  10.    len = sizeof(a)/sizeof(a[0]);
  11.    printf("%dn",len);
  12.  
  13.    for( i=0; i< len; i++)
  14.    {
  15.        val = a[i] & 0x10000000;
  16.        printf("%dn",val);
  17.        if (val > 0)
  18.           nneg++;
  19.        else
  20.           npos++;
  21.    }
  22.  
  23.    printf("npos=%d  nneg=%dn", npos, nneg);
  24.  
  25.   }
  26.  
  27.  

  
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asdasda

Answered On : Aug 10th, 2013

Code
  1. #include<conio.h>
  2. #include<iostream>
  3.  
  4. using namespace std;
  5.  
  6. int value(int a);
  7.  
  8. int main ()
  9. {
  10.     int a, num, count=0;
  11.    
  12.      cout<<"How many values: ";
  13.      cin>>num;
  14.    
  15.     getch ();
  16.    
  17.    
  18.    
  19. int value(int a);
  20.    
  21.    
  22.     for (a=1;a<=num;a++)
  23.     {
  24.    
  25.         cout<<"Enter value "<<a<<": ";
  26.         cout<<endl;
  27.         }
  28.        
  29.        
  30.         getch ();
  31.         return 0;
  32.         }
  33.        
  34.  

  
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shweta gupta

Answered On : Jun 25th, 2014

View all answers by shweta gupta

Code
  1. #include<iostream>
  2. #include<stdio.h>
  3. using namespace std;
  4. int main(){
  5. cout<<"enter the size of array"<<endl;
  6. int size,count_neg=0,count_pos=0;
  7. cin>>size;
  8. int a[size];
  9. for(int i=0;i<size;i++){
  10. cout<<"enter element of a["<<i<<"]=";
  11. cin>>a[i];
  12. if((a[i]>>31)==-1) count_neg++;
  13. else count_pos++;
  14. }
  15. cout<<"total negative numbers "<<count_neg<<" and positive number is "<<count_pos;
  16.  
  17. return 0;
  18. }

  
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