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#include stdio.h>
#include conio.h>
#include string.h>
void main()
{
int i=0,j=0,k=0,l=0;
int a[3][3],temp[3][3];
clrscr();
for(i=0;i<3;i++)
for(j=0;j<3;j++)
a[i][j]=0;
temp[i][j]=0;
}
} i=0;j=0;
printf("ENTER THE VALUE %d %d: ",i+1,j+1);
scanf("%d",&a[i][j]);
for(k=0;k<3;k++)
for(l=0;l<3;l++)
if(a[i][j] { temp[i][j]=a[i][j]; a[i][j]=a[k][l]; a[k][l]=temp[i][j]; } } } } } printf("\nNO ARE IN ACCENDING ORDER"); printf("\n"); for(i=0;i<3;i++) { for(j=0;j<3;j++) printf("%d\t",a[i][j]); printf("\n"); } getch(); } OUTPUT : ENTER THE LIMIT OF ELEMENTS: 3 ENTER THE NO 1 : 2 ENTER THE NO 2 : 3 ENTER THE NO 3 : 4 THE EVEN NO.S ARE 2 THE ODD NO.S ARE 1 THE TOTAL OF EVEN NO IS 6 THE TOTAL OF ODD NO IS 3 THE AVG. OF ALL EVEN NO IS 3.0000 THE AVG. OF ALL ODD NO IS 3.0000
temp[i][j]=a[i][j];
a[i][j]=a[k][l];
a[k][l]=temp[i][j];
printf("\nNO ARE IN ACCENDING ORDER");
printf("\n");
printf("%d\t",a[i][j]);
getch();
OUTPUT :
ENTER THE LIMIT OF ELEMENTS: 3
ENTER THE NO 1 : 2 ENTER THE NO 2 : 3
ENTER THE NO 3 : 4
THE EVEN NO.S ARE 2 THE ODD NO.S ARE 1
THE TOTAL OF EVEN NO IS 6
THE TOTAL OF ODD NO IS 3
THE AVG. OF ALL EVEN NO IS 3.0000
THE AVG. OF ALL ODD NO IS 3.0000
coding
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