Answer / azad sable, chiplun

void main()
{
int i,j,k,t,f1,f2,f3,z,sp;
sp=20;
clrscr();
for(i=0;i<=4;i++)
{
for(k=0;k<sp-1;k++)
printf("");
sp=0;
for(j=0;j<=i;j++)
{
f1=f2=f3=1;
t=i;
while(t!=0)
{
f1=f1*t;
t--;
}

t=j;
while(t!=0)
{
f2=f2*t;
t--;
}
t=i-j;

while(t!=0)
{
f3=f3*t;
t--;
}
z=f1/(f2*f3);
printf("%4d",z);
}
printf("\n");
}
getch():
}

Answer / sevak.yatrik777

with Ada.Integer_Text_Io; use Ada.Integer_Text_Io;
with Ada.Text_Io; use Ada.Text_Io;

procedure Pascals_Triangle is
type Row is array(Positive range <>) of Integer;
type Row_Access is access Row;
type Triangle is array(Positive range <>) of Row_Access;
function General_Triangle(Depth : Positive) return
Triangle is
Result : Triangle(1..Depth);
begin
for I in Result'range loop
Result(I) := new Row(1..I);
for J in 1..I loop
if J = Result(I)'First or else J =
Result(I)'Last then
Result(I)(J) := 1;
else
Result(I)(J) := Result(I - 1)(J - 1) +
Result(I - 1)(J);
end if;
end loop;
end loop;
return Result;
end General_Triangle;
procedure Print(Item : Triangle) is
begin
for I in Item'range loop
for J in 1..I loop
Put(Item => Item(I)(J), Width => 3);
end loop;
New_Line;
end loop;
end Print;
begin
Print(General_Triangle(7));
end Pascals_Triangle;

in multiple branching construct "default" case is a) optional b) compulsarily c) it is not include in this construct d) none of the above

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m=++i&&++j(||)k++ printf("%d"i,j,k,m)

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Here is a neat trick for checking whether two strings are equal

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