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Broadcasting stations need to schedule the ads, so no collision occurs. How does an ads is scheduled? Below you can find the basic.
The Algorithm 
Pascal Source Code program ad;
uses crt;
var
a:array[1..100,1..100]of integer;
c,d:array[1..100]of integer;
i,n,j,t,k,l,yedek,byk,v,old:integer;
begin
clrscr;
writeln('number of advertisements');
readln(n);
for i:=1 to n do
begin
writeln(i,' advertisement...:');
write('cost...:');readln(a[1,i]);
write('validity time...:');readln(a[2,i]);
end;
for i:=1 to n-1 do
begin
for j:=1 to n-i do
if(a[2,j]>=a[2,j+1]) then
for k:=1 to 2 do
begin
yedek:=a[k,j];
a[k,j]:=a[k,j+1];
a[k,j+1]:=yedek;
end;
end;
writeln('sorted array...:');
write('cost...: ');
for i:=1 to n do write(a[1,i]:4);
writeln;
write('validity..:');
for i:=1to n do write(a[2,i]:4);readkey; writeln;
c[1]:=a[2,1];
t:=1;
for i:=2 to n do
begin
if (a[2,i]<>a[2,i-1]) then
begin
t:=t+1;c[t]:=a[2,i];
end;
end;
for j:=t downto 1 do
begin
i:=n;
while (a[2,i]>=c[j]) and (i>1) do
begin
i:=i-1; l:=i;
end;
byk:=a[1,l];
old:=l;
for v:=l to n do
begin
if a[1,v]>byk then
begin
byk:=a[1,v];
old:=v;
end;
end;
d[j]:=byk;
a[1,old]:=0;
end;
for i:=0 to t+1 do begin
write(d[i],' ');writeln;end; readkey;
end.
Sample Execution number of advertisements 5 1. advertisement...: cost...:30 validity...:4 2. advertisement...: cost...:44 validity...:3 3. advertisement...: cost...:45 validity...:2 4. advertisement...: cost...:50 validity...:5 5. advertisement...: cost...:60 validity...:8 sorted array...: cost...: 45 44 30 50 60 validity..: 2 3 4 5 8 1. advertisement...:30 2. advertisement...:45 3. advertisement...:44 4. advertisement...:50 5. advertisement...:60
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