Solved

maximum value(location) in array

Posted on 2006-11-09
5
252 Views
Last Modified: 2010-04-15
hi,
I have a program like below how can I modify it to let it return me the location of maximum value

for example:
row 0 has maximum 4 and loacted in t[0][0]
row 0 has maximum 4 and loacted in t[0][1]
row 1 has maximum 8 and loacted in t[1][0]
row 2 has maximum 4 and loacted in t[2][2]


----------code------------------------
#include <stdio.h>
#include <stdlib.h>
#include <float.h>

                                 
int t [3][3]={{4,4,2},
              {8,2,1},
              {3,3,4}
              };                    
                   
main (){
   
   int i,j, max, allmax;
   allmax=0;
   
   for (i=0; i<3 ; i++)
   {
    max = 0;
      for (j=0; j<3; j++)
      {
        if (t[i][j]>max)
        {
          max=t[i][j];
         }
         
        if (t[i][j]>allmax)
        {
          allmax=t[i][j];
        }
      }
     
      printf("row[%d] has maximun %d and loacte in t[][]: \n",i,max );
   }

printf("allmax is : %d\n", allmax );
system("pause");
return 0;

}
0
Comment
Question by:rmtogether
  • 3
5 Comments
 
LVL 86

Expert Comment

by:jkr
ID: 17908150
Similar to your last Q, you are almost there - make that

#include <stdio.h>
#include <stdlib.h>
#include <float.h>
                                 
int t [3][3]={  {3,3,4},
                {8,2,1},
                {3,3,4}
              };                    
                   

main (){
   
   int i,j, max, allmax;

   allmax = 0;
   for (i=0; i<3 ; i++){
      max = 0;
      for (j=0; j<3; j++){
        if (t[i][j] > max) {
          max=t[i][j];
        }
        if (t[i][j] > allmax) {
          allmax=t[i][j];
        }
      }
      printf("max vaule is : %d\n",max );
   }

   printf("max vaule overall is : %d\n",allmax );


system("pause");
return 0;

}
0
 
LVL 86

Expert Comment

by:jkr
ID: 17908169
Sorry, missed the 'location part:

#include <stdio.h>
#include <stdlib.h>
#include <float.h>
                                 
int t [3][3]={  {3,3,4},
                {8,2,1},
                {3,3,4}
              };                    
                   

main (){
   
   int i,j, max, allmax, alli = 0, allj = 0;

   allmax = 0;
   for (i=0; i<3 ; i++){
      max = 0;
      for (j=0; j<3; j++){
        if (t[i][j] > max) {
          max=t[i][j];
        }
        if (t[i][j] > allmax) {
          allmax=t[i][j];
          alli = i;
          allj = j;
        }
      }
      printf("max vaule is : %d\n",max );
   }

   printf("max vaule overall is : %d at %d/%d\n",allmax, alli, allj );


system("pause");
return 0;

}
0
 

Author Comment

by:rmtogether
ID: 17908261

how can I get the maximum value for each row as well
the final output look like


row 0 has maximum 4 and loacted in t[0][0]
row 0 has maximum 4 and loacted in t[0][1]
row 1 has maximum 8 and loacted in t[1][0]
row 2 has maximum 4 and loacted in t[2][2]

max over over is 8 is in t[1][0]
0
 
LVL 86

Accepted Solution

by:
jkr earned 500 total points
ID: 17908289
Just add that part from your code again, e.g.

#include <stdio.h>
#include <stdlib.h>
#include <float.h>
                                 
int t [3][3]={  {3,3,4},
                {8,2,1},
                {3,3,4}
              };                    
                   

main (){
   
   int i,j, max, allmax, alli = 0, allj = 0, rowj = 0;

   allmax = 0;
   for (i=0; i<3 ; i++){
      max = 0;
      for (j=0; j<3; j++){
        if (t[i][j] > max) {
          max=t[i][j];
          rowj = j;
        }
        if (t[i][j] > allmax) {
          allmax=t[i][j];
          alli = i;
          allj = j;
        }
      }
      printf("max vaule is : %d at %d\n",max,rowj );
   }

   printf("max vaule overall is : %d at %d/%d\n",allmax, alli, allj );


system("pause");
return 0;

}
0
 
LVL 84

Expert Comment

by:ozo
ID: 17910172
#include <stdio.h>
#include <stdlib.h>
#include <float.h>
                                 
int t [3][3]={  {4,4,2},
                {8,2,1},
                {3,3,4}
              };                    
                   

main (){
   
   int i,j, max, allmax, alli = 0, allj = 0, rowj = 0;

   allmax = 0;
   for (i=0; i<3 ; i++){
      max = 0;
      for (j=0; j<3; j++){
        if (t[i][j] > max) {
          max=t[i][j];
          rowj = j;
        }
        if (t[i][j] > allmax) {
          allmax=t[i][j];
          alli = i;
          allj = j;
        }
      }
      for (j=0; j<3; j++){
        if( t[i][j] == max ){
         printf("row %d has maximum %d and loacted in t[%d][%d]\n",i,max,i,j );
        }
      }
   }

   printf("max vaule overall is : %d at %d/%d\n",allmax, alli, allj );


system("pause");
return 0;

}
0

Featured Post

Netscaler Common Configuration How To guides

If you use NetScaler you will want to see these guides. The NetScaler How To Guides show administrators how to get NetScaler up and configured by providing instructions for common scenarios and some not so common ones.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

Have you thought about creating an iPhone application (app), but didn't even know where to get started? Here's how: ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ Important pre-programming comments: I’ve never tri…
This is a short and sweet, but (hopefully) to the point article. There seems to be some fundamental misunderstanding about the function prototype for the "main" function in C and C++, more specifically what type this function should return. I see so…
The goal of this video is to provide viewers with basic examples to understand and use structures in the C programming language.
Video by: Grant
The goal of this video is to provide viewers with basic examples to understand and use for-loops in the C programming language.

910 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question

Need Help in Real-Time?

Connect with top rated Experts

20 Experts available now in Live!

Get 1:1 Help Now