#include<cstdio>
int main() {
int n,i,ul,a,x,y,z,s=0; //n is number of tiles, i is counter variable , ul is upper limit & a,x,y,z are temp variables while s is sum variable for counting all possibilites
scanf("%d",&n);
ul=(int)n/2;
for (i=1;i<=ul;i++) {
x=1;y=1;z=1;
for (a=(n-i);a>1;a--) {if ((n-i)==1) break;x=x*a;} //calculating (n-i) factorial
for (a=i;a>1;a--) {if ((i)==1) break;y=y*a;} //calculating i factorial
for (a=(n-(2*i));a>=1;a--) {if ((n-(2*i))==1) break;z=(int)z*a;} //calculating (n-(2*i)) factorial
s+=(int)x/(y*z);
}
printf("%d",(s+1));
}
I am again getting a Floating Point exception , i know what is this error , but cant remove it from my code, can you pls help ???
For further reference here's the Question to which it is solution :-
To help Lavanya learn all about binary numbers and binary sequences, her father has bought her a collection of square tiles, each of which has either a 0 or a 1 written on it. Her brother Nikhil has played a rather nasty prank. He has glued together pairs of tiles with 0 written on them. Lavanya now has square tiles with 1 on them and rectangular tiles with two 0's on them, made up of two square tiles with 0 stuck together). Thus, she can no longer make all possible binary sequences using these tiles.
To amuse herself, Lavanya has decided to pick a number N and try and construct as many binary sequences of length N as possible using her collection of tiles. For example if N = 1, she can only make the sequence 1. For N=2, she can make 11 and 00. For N=4, there are 5 possiblities: 0011, 0000, 1001, 1100 and 1111.
Lavanya would like you to write a program to compute the number of arrangements possible with N tiles so that she can verify that she has generated all of them. Since she cannot count beyond 15746, it is sufficient to report this number modulo 15746.
Input format
A single line with a single integer N.
Output format
A single integer indicating the number of binary sequences of length N, modulo 15746, that Lavanya can make using her tiles.
Test data
You may assume that N ≤ 1000000.
Example
Here is a sample input and output corresponding to the example discussed above.
Yeah you pointed out correct possibility and i edited my code accordingly but still it gives an "Floating Point Error" , here's edited code edited on Line 12 :-
1 2 3 4 5 6 7 8 9 10 11 12 13 14
#include<cstdio>
int main() {
int n,i,ul,a,x,y,z,s=0;
scanf("%d",&n);
ul=(int)n/2;
for (i=1;i<=ul;i++) {
x=1;y=1;z=1;
for (a=(n-i);a>1;a--) {if ((n-i)==1) break;x=x*a;}
for (a=i;a>1;a--) {if ((i)==1) break;y=y*a;}
for (a=(n-(2*i));a>=1;a--) {if ((n-(2*i))==1) break;z=(int)z*a;}
if ((y*z)!=0) s+=(int)x/(y*z); //condition here is to avoid division by 0
}
printf("%d",(s+1));
}