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Join Date: Nov 2007

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nurulshidanoni nurulshidanoni is offline

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min value

Nov 6th, 2007

hello,
why the output below produce number 0, eventhought there is no 0 in the array?

 Help with Code Tags 
 c++ Syntax (Toggle Plain Text) 
#include <iostream>
#include <conio.h>
#include <cstdlib>
#include <cmath>
#include <cstdio>
#include <ctime>
using namespace std;
 
int i;
int x[7];
int A[7], B[7], C[7], D[7], E[7], F[7];
int Min( const int *A, const int Count);
int minimum, minA, minB, flagB;
int main(void);
int count = 7;
{
    for(i=1;i<7;i++)
{
       x[i]=i;
       A[i]=(2+pow(x[i],2));
       B[i]=(1+pow(x[i],3));
       C[i]=(2+pow(x[i],2)-1);
       D[i]=(3*pow(x[i],3));
       E[i]=(1+pow(x[i],2));
       F[i]=(3+pow(x[i],2));
}
{
       cout <<"\n";
 
       for(i=1;i<7; i++)
{      printf("[%d][%d] A[%d]=%d\n", i,1, i, A[i]);
}
       cout <<"\n";
       for(i=1;i<7; i++)
{
       printf("[%d][%d] B[%d]=%d\n", i,2, i, B[i]);
}
       cout <<"\n";
       for(i=1;i<7; i++)
{
       printf("[%d][%d] C[%d]=%d\n", i,3, i, C[i]);
}
       cout <<"\n";
       for(i=1;i<7; i++)
{
       printf("[%d][%d] D[%d]=%d\n", i,4, i, D[i]);
 
}
       cout <<"\n";
       for(i=1;i<7; i++)
{
       printf("[%d][%d] E[%d]=%d\n", i,5, i, E[i]);
}
       cout <<"\n";
       for(i=1;i<7; i++)
{
       printf("[%d][%d] F[%d]=%d\n", i,6, i, F[i]);
}
}
       printf("\n");
       printf ("A\tB\tC\tD\tE\tF");
       printf ("\n_\t_\t_\t_\t_\t_\t");
       for(i=1;i<7;i++)
{
       cout<<endl;
       printf("%d\t", A[i]);
       printf("%d\t", B[i]);
       printf("%d\t", C[i]);
       printf("%d\t", D[i]);
       printf("%d\t", E[i]);
       printf("%d\t", F[i]);
}
       cout <<"\n";
    // Compare the members
       int Minimum = A[0];
	   for (int i=1; i<7; i++)
		   if (minimum > A[i]) Minimum = A[i];
		   return minimum;
}
 
    // Announce the result
       cout << "The minimum value of the array A is "<< A[minA] << "." <<   endl;
}
	   }
#include <iostream>
#include <conio.h>
#include <cstdlib>
#include <cmath>
#include <cstdio>
#include <ctime>
using namespace std;

int i;
int x[7];
int A[7], B[7], C[7], D[7], E[7], F[7];
int Min( const int *A, const int Count);
int minimum, minA, minB, flagB;
int main(void);
int count = 7;
{
    for(i=1;i<7;i++)
{
       x[i]=i;
       A[i]=(2+pow(x[i],2));
       B[i]=(1+pow(x[i],3));
       C[i]=(2+pow(x[i],2)-1);
       D[i]=(3*pow(x[i],3));
       E[i]=(1+pow(x[i],2));
       F[i]=(3+pow(x[i],2));
}
{
       cout <<"\n";

       for(i=1;i<7; i++)
{      printf("[%d][%d] A[%d]=%d\n", i,1, i, A[i]);
}
       cout <<"\n";
       for(i=1;i<7; i++)
{
       printf("[%d][%d] B[%d]=%d\n", i,2, i, B[i]);
}
       cout <<"\n";
       for(i=1;i<7; i++)
{
       printf("[%d][%d] C[%d]=%d\n", i,3, i, C[i]);
}
       cout <<"\n";
       for(i=1;i<7; i++)
{
       printf("[%d][%d] D[%d]=%d\n", i,4, i, D[i]);

}
       cout <<"\n";
       for(i=1;i<7; i++)
{
       printf("[%d][%d] E[%d]=%d\n", i,5, i, E[i]);
}
       cout <<"\n";
       for(i=1;i<7; i++)
{
       printf("[%d][%d] F[%d]=%d\n", i,6, i, F[i]);
}
}
       printf("\n");
       printf ("A\tB\tC\tD\tE\tF");
       printf ("\n_\t_\t_\t_\t_\t_\t");
       for(i=1;i<7;i++)
{
       cout<<endl;
       printf("%d\t", A[i]);
       printf("%d\t", B[i]);
       printf("%d\t", C[i]);
       printf("%d\t", D[i]);
       printf("%d\t", E[i]);
       printf("%d\t", F[i]);
}
       cout <<"\n";
    // Compare the members
       int Minimum = A[0];
	   for (int i=1; i<7; i++)
		   if (minimum > A[i]) Minimum = A[i];
		   return minimum;
}

    // Announce the result
       cout << "The minimum value of the array A is "<< A[minA] << "." <<   endl;
}
	   }

Last edited by Ancient Dragon; Nov 6th, 2007 at 9:26 pm. Reason: add code tags

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n.aggel

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Re: min value

Nov 6th, 2007

please use code tags or other people won't be able to help you....

here is your code

 Help with Code Tags 
 c++ Syntax (Toggle Plain Text) 
#include <iostream>
#include <conio.h>
#include <cstdlib>
#include <cmath>
#include <cstdio>
#include <ctime>
using namespace std;
 
int i;
int x[7];
int A[7], B[7], C[7], D[7], E[7], F[7];
int Min( const int *A, const int Count);
int minimum, minA, minB, flagB;
int main(void);
int count = 7;
{
    for(i=1;i<7;i++)
{
       x[i]=i;
       A[i]=(2+pow(x[i],2));
       B[i]=(1+pow(x[i],3));
       C[i]=(2+pow(x[i],2)-1);
       D[i]=(3*pow(x[i],3));
       E[i]=(1+pow(x[i],2));
       F[i]=(3+pow(x[i],2));
}
{
       cout <<"\n";
 
       for(i=1;i<7; i++)
{      printf("[%d][%d] A[%d]=%d\n", i,1, i, A[i]);
}
       cout <<"\n";
       for(i=1;i<7; i++)
{
       printf("[%d][%d] B[%d]=%d\n", i,2, i, B[i]);
}
       cout <<"\n";
       for(i=1;i<7; i++)
{
       printf("[%d][%d] C[%d]=%d\n", i,3, i, C[i]);
}
       cout <<"\n";
       for(i=1;i<7; i++)
{
       printf("[%d][%d] D[%d]=%d\n", i,4, i, D[i]);
 
}
       cout <<"\n";
       for(i=1;i<7; i++)
{
       printf("[%d][%d] E[%d]=%d\n", i,5, i, E[i]);
}
       cout <<"\n";
       for(i=1;i<7; i++)
{
       printf("[%d][%d] F[%d]=%d\n", i,6, i, F[i]);
}
}
       printf("\n");
       printf ("A\tB\tC\tD\tE\tF");
       printf ("\n_\t_\t_\t_\t_\t_\t");
       for(i=1;i<7;i++)
{
       cout<<endl;
       printf("%d\t", A[i]);
       printf("%d\t", B[i]);
       printf("%d\t", C[i]);
       printf("%d\t", D[i]);
       printf("%d\t", E[i]);
       printf("%d\t", F[i]);
}
       cout <<"\n";
    // Compare the members
       int Minimum = A[0];
	   for (int i=1; i<7; i++)
		   if (minimum > A[i]) Minimum = A[i];
		   return minimum;
}
 
    // Announce the result
       cout << "The minimum value of the array A is "<< A[minA] << "." <<   endl;
}
	   }
#include <iostream>
#include <conio.h>
#include <cstdlib>
#include <cmath>
#include <cstdio>
#include <ctime>
using namespace std;

int i;
int x[7];
int A[7], B[7], C[7], D[7], E[7], F[7];
int Min( const int *A, const int Count);
int minimum, minA, minB, flagB;
int main(void);
int count = 7;
{
    for(i=1;i<7;i++)
{
       x[i]=i;
       A[i]=(2+pow(x[i],2));
       B[i]=(1+pow(x[i],3));
       C[i]=(2+pow(x[i],2)-1);
       D[i]=(3*pow(x[i],3));
       E[i]=(1+pow(x[i],2));
       F[i]=(3+pow(x[i],2));
}
{
       cout <<"\n";

       for(i=1;i<7; i++)
{      printf("[%d][%d] A[%d]=%d\n", i,1, i, A[i]);
}
       cout <<"\n";
       for(i=1;i<7; i++)
{
       printf("[%d][%d] B[%d]=%d\n", i,2, i, B[i]);
}
       cout <<"\n";
       for(i=1;i<7; i++)
{
       printf("[%d][%d] C[%d]=%d\n", i,3, i, C[i]);
}
       cout <<"\n";
       for(i=1;i<7; i++)
{
       printf("[%d][%d] D[%d]=%d\n", i,4, i, D[i]);

}
       cout <<"\n";
       for(i=1;i<7; i++)
{
       printf("[%d][%d] E[%d]=%d\n", i,5, i, E[i]);
}
       cout <<"\n";
       for(i=1;i<7; i++)
{
       printf("[%d][%d] F[%d]=%d\n", i,6, i, F[i]);
}
}
       printf("\n");
       printf ("A\tB\tC\tD\tE\tF");
       printf ("\n_\t_\t_\t_\t_\t_\t");
       for(i=1;i<7;i++)
{
       cout<<endl;
       printf("%d\t", A[i]);
       printf("%d\t", B[i]);
       printf("%d\t", C[i]);
       printf("%d\t", D[i]);
       printf("%d\t", E[i]);
       printf("%d\t", F[i]);
}
       cout <<"\n";
    // Compare the members
       int Minimum = A[0];
	   for (int i=1; i<7; i++)
		   if (minimum > A[i]) Minimum = A[i];
		   return minimum;
}

    // Announce the result
       cout << "The minimum value of the array A is "<< A[minA] << "." <<   endl;
}
	   }

what compiler do you use? on gcc it doesn't compile....

remove line 2:: #include <conio.h>

also in lines like A[i]=(2+pow(x[i],2)); you must cast the operands to double, so that there is no amguity as to which overloaded function of pow the compiler selects..

to do this type:: A[i]=( 2+pow( static_cast<double>(x[i]),static_cast<double>(2) ) );

Two roads diverged in a wood, and I— I took the one less traveled by, and that has made all the difference.

by Robert Frost the "The Road Not Taken"

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Join Date: Nov 2007

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nurulshidanoni nurulshidanoni is offline

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Re: min value

Nov 6th, 2007

How to use code tags?
whata is complier?
I use visual basic c++ 6.0 edition, is that compiler?
Thanks for the correction, but why A[minA], doesnt produce the value that I want?

cout << "The minimum value of the array A is "<< A[minA] << "." << endl;

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Re: min value

Nov 6th, 2007

>>How to use code tags?
Look in the edit box. The instructions are in grey letters. All you have to do is read them.

>>whata is complier?
Too bad I can't issue an infraction for being such a dumbass.

But I hope you were just teasing.

lines 20-25 of your original post: array x is initialized by the program's start-up code to be 0 because the array is in global memory. So 0 ^ <any number here> is always 0.

Last edited by Ancient Dragon; Nov 6th, 2007 at 9:30 pm.

Don't PM me with questions -- you might get a nasty PM in response. If you have a question then post it in one of the forums.

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nurulshidanoni nurulshidanoni is offline

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Re: min value

Nov 6th, 2007

i have get the minimum value that I want, but how to get the min from which array?I have done in bold, but its seems like wrong

 Help with Code Tags 
 c++ Syntax (Toggle Plain Text) 
#include <iostream>
#include <cstdlib>
#include <cmath>
#include <cstdio>
#include <ctime>
using namespace std;
 
int i;
int x[7];
int A[7], B[7], C[7], D[7], E[7], F[7];
// The members of the array int minimum = numbers[0] ;
int minimum, minA, minB, flagB;
int main(void)
{
    for(i=1;i<7;i++)
{
       x[i]=i;
       A[i]=(2+pow(static_cast<double>(x[i]),static_cast<double>(2)));
       B[i]=(1+pow(static_cast<double>(x[i]),static_cast<double>(3)));
       C[i]=(2+pow(static_cast<double>(x[i]),static_cast<double>(2)-1));
       D[i]=(3*pow(static_cast<double>(x[i]),static_cast<double>(3)));
       E[i]=(1+pow(static_cast<double>(x[i]),static_cast<double>(2)));
       F[i]=(3+pow(static_cast<double>(x[i]),static_cast<double>(2)));
}
{
       cout <<"\n";
 
       for(i=1;i<7; i++)
{      printf("[%d][%d] A[%d]=%d\n", i,1, i, A[i]);
}
       cout <<"\n";
       for(i=1;i<7; i++)
{
       printf("[%d][%d] B[%d]=%d\n", i,2, i, B[i]);
}
       cout <<"\n";
       for(i=1;i<7; i++)
{
       printf("[%d][%d] C[%d]=%d\n", i,3, i, C[i]);
}
       cout <<"\n";
       for(i=1;i<7; i++)
{
       printf("[%d][%d] D[%d]=%d\n", i,4, i, D[i]);
 
}
       cout <<"\n";
       for(i=1;i<7; i++)
{
       printf("[%d][%d] E[%d]=%d\n", i,5, i, E[i]);
}
       cout <<"\n";
       for(i=1;i<7; i++)
{
       printf("[%d][%d] F[%d]=%d\n", i,6, i, F[i]);
}
}
       printf("\n");
       printf ("A\tB\tC\tD\tE\tF");
       printf ("\n_\t_\t_\t_\t_\t_\t");
       for(i=1;i<7;i++)
{
       cout<<endl;
       printf("%d\t", A[i]);
       printf("%d\t", B[i]);
       printf("%d\t", C[i]);
       printf("%d\t", D[i]);
       printf("%d\t", E[i]);
       printf("%d\t", F[i]);
}
       cout <<"\n";
    // Compare the members
       int minimum = A[1];
       for (i = 1; i < 7; ++i)
{
       if ( A[i] < minimum) 
{
 
	   minimum=A[i];
	   minA = i;
	   return minimum;
 
}      cout <<"\n";
    // Announce the result
       cout << "The minimum value of the array A is "<< minimum << "." <<   endl;
 
       minimum = B[1];
       for (i = 1; i < 7; ++i)
{
       if (B[i]< minimum)
{
		minimum = B[i];
    	minB = i; 
}
        cout <<"\n";
        cout << "The minimum value of the arrays B is "<< minimum << "." << endl;
 
   // to determine the minimum whether in A or B
	    minimum = A[minA];
	    if(B[minB] < minimum)
{
		minimum = B[minB];
		flagB = 1; //the minimum is in B if flagB=1
}
    	cout <<"\n";
        [B]if(flagB = 1)
		cout<< "Initial solution in A:" <<minimum << "."<<endl;
	    else 
		cout<< "Initial solution in B:" <<minimum << "."<<endl;
 
        return 0;[/B]}
}
}
#include <iostream>
#include <cstdlib>
#include <cmath>
#include <cstdio>
#include <ctime>
using namespace std;

int i;
int x[7];
int A[7], B[7], C[7], D[7], E[7], F[7];
// The members of the array int minimum = numbers[0] ;
int minimum, minA, minB, flagB;
int main(void)
{
    for(i=1;i<7;i++)
{
       x[i]=i;
       A[i]=(2+pow(static_cast<double>(x[i]),static_cast<double>(2)));
       B[i]=(1+pow(static_cast<double>(x[i]),static_cast<double>(3)));
       C[i]=(2+pow(static_cast<double>(x[i]),static_cast<double>(2)-1));
       D[i]=(3*pow(static_cast<double>(x[i]),static_cast<double>(3)));
       E[i]=(1+pow(static_cast<double>(x[i]),static_cast<double>(2)));
       F[i]=(3+pow(static_cast<double>(x[i]),static_cast<double>(2)));
}
{
       cout <<"\n";

       for(i=1;i<7; i++)
{      printf("[%d][%d] A[%d]=%d\n", i,1, i, A[i]);
}
       cout <<"\n";
       for(i=1;i<7; i++)
{
       printf("[%d][%d] B[%d]=%d\n", i,2, i, B[i]);
}
       cout <<"\n";
       for(i=1;i<7; i++)
{
       printf("[%d][%d] C[%d]=%d\n", i,3, i, C[i]);
}
       cout <<"\n";
       for(i=1;i<7; i++)
{
       printf("[%d][%d] D[%d]=%d\n", i,4, i, D[i]);

}
       cout <<"\n";
       for(i=1;i<7; i++)
{
       printf("[%d][%d] E[%d]=%d\n", i,5, i, E[i]);
}
       cout <<"\n";
       for(i=1;i<7; i++)
{
       printf("[%d][%d] F[%d]=%d\n", i,6, i, F[i]);
}
}
       printf("\n");
       printf ("A\tB\tC\tD\tE\tF");
       printf ("\n_\t_\t_\t_\t_\t_\t");
       for(i=1;i<7;i++)
{
       cout<<endl;
       printf("%d\t", A[i]);
       printf("%d\t", B[i]);
       printf("%d\t", C[i]);
       printf("%d\t", D[i]);
       printf("%d\t", E[i]);
       printf("%d\t", F[i]);
}
       cout <<"\n";
    // Compare the members
       int minimum = A[1];
       for (i = 1; i < 7; ++i)
{
       if ( A[i] < minimum) 
{
    
	   minimum=A[i];
	   minA = i;
	   return minimum;

}      cout <<"\n";
    // Announce the result
       cout << "The minimum value of the array A is "<< minimum << "." <<   endl;

       minimum = B[1];
       for (i = 1; i < 7; ++i)
{
       if (B[i]< minimum)
{
		minimum = B[i];
    	minB = i; 
}
        cout <<"\n";
        cout << "The minimum value of the arrays B is "<< minimum << "." << endl;
 
   // to determine the minimum whether in A or B
	    minimum = A[minA];
	    if(B[minB] < minimum)
{
		minimum = B[minB];
		flagB = 1; //the minimum is in B if flagB=1
}
    	cout <<"\n";
        [B]if(flagB = 1)
		cout<< "Initial solution in A:" <<minimum << "."<<endl;
	    else 
		cout<< "Initial solution in B:" <<minimum << "."<<endl;
	   
        return 0;[/B]}
}
}

Last edited by Ancient Dragon; Nov 6th, 2007 at 10:49 pm. Reason: replace quote tags with code tags

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Re: min value

Nov 6th, 2007

did you compile that? Probably not because it has lots of errors. Start out by counting the open and close braced and make sure they match correctly. Next fix the indention -- its absolutely horrible. Don't be afraid to use the space bar to align the braces and code.

lines 18-23. What is the purpose of X array? Why do you even need it? Did you read what I previously posted about it? Obviously not because you didn't change it.

Don't PM me with questions -- you might get a nasty PM in response. If you have a question then post it in one of the forums.

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Re: min value

Nov 6th, 2007

What is the purpose of X array?
the pupose is to get the value.

•

[1][1] A[1]=3
[2][1] A[2]=6
[3][1] A[3]=11
[4][1] A[4]=18
[5][1] A[5]=27
[6][1] A[6]=38

[1][2] B[1]=2
[2][2] B[2]=9
[3][2] B[3]=28
[4][2] B[4]=65
[5][2] B[5]=126
[6][2] B[6]=217

[1][3] C[1]=3
[2][3] C[2]=4
[3][3] C[3]=5
[4][3] C[4]=6
[5][3] C[5]=7
[6][3] C[6]=8

[1][4] D[1]=3
[2][4] D[2]=24
[3][4] D[3]=81
[4][4] D[4]=192
[5][4] D[5]=375
[6][4] D[6]=648

[1][5] E[1]=2
[2][5] E[2]=5
[3][5] E[3]=10
[4][5] E[4]=17
[5][5] E[5]=26
[6][5] E[6]=37

[1][6] F[1]=4
[2][6] F[2]=7
[3][6] F[3]=12
[4][6] F[4]=19
[5][6] F[5]=28
[6][6] F[6]=39

A B C D E F
_ _ _ _ _ _
3 2 3 3 2 4
6 9 4 24 5 7
11 28 5 81 10 12
18 65 6 192 17 19
27 126 7 375 26 28
38 217 8 648 37 39

The minimum value of the array A is 3.

The minimum value of the arrays B is 2.

Initial solution in B: 0.
Press any key to continue

I have compile bit for the initial solution, why is it 0?

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Re: min value

Nov 6th, 2007

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Originally Posted by nurulshidanoni

What is the purpose of X array?
the pupose is to get the value.

To get the value of what? All elements of that array are initialized to 0 and you never change them.

•

Originally Posted by nurulshidanoni

I have compile bit for the initial solution, why is it 0?

See above.

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Re: min value

Nov 6th, 2007

Initialize to zero,? Dont you see that the array gives the value. Its a mathematic eqyuation, and i arrange in a matrix array.

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Re: min value

#10

Nov 6th, 2007

Yes I see now its initialized on line 19. But still, why do you need that array. You can just simply do this:

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for(i=1;i<7;i++)
{
       A[i]=(2+pow(i, 2.0);
       B[i]=(1+pow(i,3.0);
       C[i]=(2+pow(i,1.0);
       D[i]=(3*pow(i,3.0);
       E[i]=(1+pow(I,2.0);
       F[i]=(3+pow(I,2.0);
}
for(i=1;i<7;i++)
{
       A[i]=(2+pow(i, 2.0);
       B[i]=(1+pow(i,3.0);
       C[i]=(2+pow(i,1.0);
       D[i]=(3*pow(i,3.0);
       E[i]=(1+pow(I,2.0);
       F[i]=(3+pow(I,2.0);
}