since your input is of fixed size, your algorithm always runs in
constant time. If the # of students and # of courses are variable, the
algorithm is O(n^2).
satya.
On 5/28/07, sl7fat <[EMAIL PROTECTED]> wrote:
>
> hi i have an algorthim code and i have to find the time complixcity of
> the code so can you plz help me ASAP the code is written done ,,
> # include <iostream.h>
>
> void main()
> {
>
> int a[10][4]=
> {{ 16,17,19,13},
> {18,14,15,19},
> {18,20,20,19},
> {13,14,15,10},
> {20,17,19,19},
> {18,13,18,19},
> {18,10,15,12},
> {12,14,15,11},
> {12,16,17,18},
> {18,11,15,10}} ;
>
>
> int i,j,max,min;
> float avg,sum;
>
>
> for(i=0;i<10;i++)
> {
>
> for(j=0;j<4;j++)
> {
>
> cout << a[i][j] << " ";
> }
> cout << "\n";
> }
>
> for(i=0;i<4;i++)
> {
> max= a[0][i];
> min= a[0][i];
> sum=0;
>
> for(j=1;j<10;j++)
> {
>
> sum= sum+a[j][i];
> if(a[j][i]>max)
> max=a[j][i];
>
> if(a[j][i]<min)
> min=a[j][i];
> }
>
> avg= sum/10;
>
> cout << "The average Grade for Exam "<<i+1<< " is: "<<
> avg<<"\n";
> cout << "The minimum Grade for Exam "<<i+1<< " is: "<<
> min<<"\n";
> cout << "The maximum Grade for Exam "<<i+1<< " is: "<<
> max<<"\n\n";
>
>
> }
>
> for(i=0;i<10;i++)
> {
> min= a[i][0];
> sum=0;
> for(j=0;j<4;j++)
> {
> sum= sum+a[i][j];
> if ( min>a[i][j])
> min= a[i][j];
>
> }
> sum= sum-min;
>
> cout << " The summation of the best 3 grades for student No
> "<< i
> +1<< " is: "<< sum<<"\n";
> }
>
> }
>
> thanx alot :D
>
>
> >
>
--
...what's remarkable, is that atoms have assembled into entities which
are somehow able to ponder their origins.
--
http://cs.uic.edu/~spopuri
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