write a c program to print magic square of order n when n>3
and n is odd?
Answer / v p p
#include<stdio.h>
#include<conio.h>
void main()
{
int i,j=1,k,n,a[22][22],p,q,r,count=0;
clrscr();
A:
printf("Enter No. Of Rows Or Columns(Should Be Odd) : ");
scanf("%d",&n);
printf("\n");
if(n%2==0)
{
printf("Enter Odd No.s Only\n\n");
goto A;
}
k=(n+1)/2;
for(p=1;p<=n;p++)
{ for(q=1;q<=n;q++)
{
a[p][q]=NULL;
}
}
for(i=1;i<=(n*n);i++)
{
if(a[j][k]==NULL)
{
a[j][k]=i;
}
else
{
j=j+2;
k--;
if(j==n+2)
j=2;
if(k==0)
k=n;
if(j==0)
j=n;
if(k==n+1)
k=1;
a[j][k]=i;
}
j--;
k++;
if(j==0)
j=n;
if(k==n+1)
k=1;
}
for(p=1;p<=n;p++)
{
for(q=1;q<=n;q++)
{
r=a[p][q];
while(r>0)
{
r=r/10;
count++;
}
if(count==1)
printf(" 0%d",a[p][q]);
else
printf(" %d",a[p][q]);
if(q%n==0)
printf("\n\n");
count=0;
}
}
for(p=1;p<=n;p++)
{
for(q=1;q<=n;q++)
{
}
}
getch();
}
| Is This Answer Correct ? | 18 Yes | 6 No |
void main() { int *mptr, *cptr; mptr = (int*)malloc(sizeof(int)); printf(ā%dā,*mptr); int *cptr = (int*)calloc(sizeof(int),1); printf(ā%dā,*cptr); }
I need your help, i need a Turbo C code for this problem.. hope u'll help me guys.? Your program will have a 3x3 array. The user will input the sum of each row and each column. Then the user will input 3 values and store them anywhere, or any location or index, temporarily in the array. Your program will supply the remaining six (6) values and determine the exact location of each value in the array. Example: Input: Sum of row 1: 6 Sum of row 2: 15 Sum of row 3: 24 Sum of column 1: 12 Sum of column 2: 15 Sum of column 3: 18 Value 1: 3 Value 2: 5 Value 3: 6 Output: Sum of Row 1 2 3 6 4 5 6 15 7 8 9 24 Sum of Column 12 15 18 Note: Your program will not necessary sort the walues in the array Thanks..
#include<stdio.h> #include<conio.h> void main() { int a=(1,2,3,(1,2,3,4); switch(a) { printf("ans:"); case 1: printf("1");break; case 2: printf("2");break; case 3: printf("1");break; case 4: printf("4");break; printf("end"); } getch(); }
main() { printf("\nab"); printf("\bsi"); printf("\rha"); }
3) Int Matrix of certain size was given, We had few valu= es in it like this. =97=97=97=97=97=97=97=97=97=97=97 1 = | 4 | | 5 | &= nbsp; | 45 =97=97=97=97=97=97=97=97=97=97=97 &n= bsp; | 3 | 3 | 5 | = | 4 =97=97=97=97=97=97=97=97=97=97=97 34 |&nbs= p; 3 | 3 | | 12 | &= nbsp; =97=97=97=97=97=97=97=97=97=97=97 3 | &nbs= p; | 3 | 4 | = | 3 =97=97=97=97=97=97=97=97=97=97=97 3 | = ; | | | = ; 3 | =97=97=97=97=97=97=97=97=97=97=97 &= nbsp; | | 4 | = ; | 4 | 3 We w= ere supposed to move back all the spaces in it at the end. Note: = If implemented this prog using recursion, would get higher preference.
All the combinations of prime numbers whose sum gives 32
void main() { char a[]="12345\0"; int i=strlen(a); printf("here in 3 %d\n",++i); }
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