Discussion @ 1044. Lucky Tickets. Easy! @ Timus Online Judge

SwSwordHoly o(100000000) [4] // Problem 1044. Lucky Tickets. Easy! 8 Mar 2009 08:34

#include<stdio.h>
void main()
{

int a[10];
int i,j,n,i1,i2,i3,i4,i5,i6,i7,i8;
a[1]=10;
a[2]=10;
a[4]=0;
a[6]=0;
a[8]=0;
for(i1=0;i1<=9;i1++)
   for(i2=0;i2<=9;i2++)
     for(i3=0;i3<=9;i3++)
      for(i4=0;i4<=9;i4++)
          if ((i1+i2)==(i3+i4)) a[4]++;
for(i1=0;i1<=9;i1++)
   for(i2=0;i2<=9;i2++)
     for(i3=0;i3<=9;i3++)
      for(i4=0;i4<=9;i4++)
          for(i5=0;i5<=9;i5++)
            for(i6=0;i6<=9;i6++)
                if((i1+i2+i3)==(i4+i5+i6)) a[6]++;

for(i1=0;i1<=9;i1++)
   for(i2=0;i2<=9;i2++)
     for(i3=0;i3<=9;i3++)
      for(i4=0;i4<=9;i4++)
          for(i5=0;i5<=9;i5++)
            for(i6=0;i6<=9;i6++)
              for(i7=0;i7<=9;i7++)
                 for(i8=0;i8<=9;i8++)
                     if((i1+i2+i3+i4)==(i5+i6+i7+i8)) a[8]++;

scanf("%d",&n);
printf("%d",a[n]);

}
o(100000000) 0.218sec

Thanawat Re: o(100000000) [1] // Problem 1044. Lucky Tickets. Easy! 28 Oct 2009 13:28

can u tell me how to solve this problem.

phizaz Re: o(100000000) // Problem 1044. Lucky Tickets. Easy! 6 Apr 2011 12:52

Explaination of that solution is...
simulating number for each digits ... it's seems to "for(i=1;i<=99999999;i++)" but it's easier(faster) to check for SUM of first half and the second half.

Edited by author 06.04.2011 12:53

phizaz Re: o(100000000) // Problem 1044. Lucky Tickets. Easy! 6 Apr 2011 12:47

Thanks. you opened my eyes!
Now, I write it in recursive terms of you!

Edited by author 06.04.2011 12:58

Alisher Alisher_uz // Problem 1044. Lucky Tickets. Easy! 23 Apr 2011 23:16

thank you very much

Discussion of Problem 1044. Lucky Tickets. Easy!