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retreiving the maxnum & minnum
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Join Date: Jul 2004
Posts: 5
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can anyone explain to me a simple way to get the lowest input and the highest input From multiple inputs
while also counting how many are below a certain value and avove a certain value
I am at my witts end tryng to learn. I am a newbie at programming and trying so many ways that I know are unneeded so far I have this
#include <iostream.h>
main()
{
int totalCalls,callcounter,Duration;
int longest, shortest;
double average ;
while(callcounter <= 10) //using ten as the amount o call to easy use
{
cout << "Enter Call duration :\n"; \\ lenght of calls
cin >> Duration ;\\ user input
totalcalls = totalcalls + duration;\\ adds duration to call total
callcounter+ ; \\add one or every loop to counter
}//end while
average = totalcalls/callcounter;\\ average of all calls
if (duration ==5)
shortest = duration
cout << " Average length of all calls :"<< average <<"\n";
cout << " Total length of all calls :"<< totalcalls<<"\n";
cout << " Total of call\t :"<< callcounter<<"\n";
}endmain
I have tried multiple if statements but am lost
I know it may be possible that I have to use arrays, switchs or a pointers
but as I have just started to learn C++ within the last two weeks any help is appreciated
<
If it aint hard it it good > this function get the minimum value for numbers that the user will input
#include<iostream.h>
using namespace std;
int min(const in *Numbers, const int Count)
{
int Minimum = Numbers[0];
for(int i = 0; i < Count; i++)
Minimum = Numbers[i];
return Minimum;
}
double Min(const double *Numbers, const int Count)
{
double Minimum = Numbers[0];
for(int i = 0; i < Count; i++)
if( Minimum > Numbers[i] )
Minimum = Numbers[i];
return Minimum;
}
double Min(const double *Numbers, const int Count)
{
double Minimum = Numbers[0];
for(int i = 0; i < Count; i++)
if( Minimum > Numbers[i] )
Minimum = Numbers[i];
return Minimum;
}
int main()
{
int Nbrs[] = { 12, 483, 748, 35, 478 };
int Total = sizeof(Nbrs) / sizeof(int);
int Minimum = Min(Nbrs, Total);
cout << "Minimum: " << Minimum << endl;
return 0;
}
Here is an example of running the program:
Minimum: 12Press any key to continue
#include<iostream.h>
using namespace std;
int min(const in *Numbers, const int Count)
{
int Minimum = Numbers[0];
for(int i = 0; i < Count; i++)
Minimum = Numbers[i];
return Minimum;
}
double Min(const double *Numbers, const int Count)
{
double Minimum = Numbers[0];
for(int i = 0; i < Count; i++)
if( Minimum > Numbers[i] )
Minimum = Numbers[i];
return Minimum;
}
double Min(const double *Numbers, const int Count)
{
double Minimum = Numbers[0];
for(int i = 0; i < Count; i++)
if( Minimum > Numbers[i] )
Minimum = Numbers[i];
return Minimum;
}
int main()
{
int Nbrs[] = { 12, 483, 748, 35, 478 };
int Total = sizeof(Nbrs) / sizeof(int);
int Minimum = Min(Nbrs, Total);
cout << "Minimum: " << Minimum << endl;
return 0;
}
Here is an example of running the program:
Minimum: 12Press any key to continue
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Hmm how about declaring 2 macros:
I hope it's write since I didn't test it. :o
C++ Syntax (Toggle Plain Text)
#define max(a,b) a>b?a:b #define min(a,b) a<b?a:b int max,int min; ... max=max(x[0],x[1]); min=min(x[0],x[1]); for (i=2;i<n;i++) { max=max(max,x[i]); min=min(min,x[i]); } ...
I hope it's write since I didn't test it. :o
take a look at ..
http://www.daniweb.com/techtalkforum...7974#post37974
http://www.daniweb.com/techtalkforum...7974#post37974
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Join Date: Jul 2004
Posts: 5
Reputation:
Solved Threads: 0
whoa there now I appreciate the help
that macro was a lil bit above my head
but I will look into that also meab thanks for the post
with maxnum function
I found something that was more simple to use
[code]
for (j=0;j >= 100 ;j++)
counter++
[\code]
this way I can keep count of all my inputs and determine the max and min
I think , But thanks to all who have contributed
<if it aint hard it aint right >
that macro was a lil bit above my head
but I will look into that also meab thanks for the post
with maxnum function
I found something that was more simple to use
[code]
for (j=0;j >= 100 ;j++)
counter++
[\code]
this way I can keep count of all my inputs and determine the max and min
I think , But thanks to all who have contributed
<if it aint hard it aint right >
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