This is a relatively simple math problem: the number of chickens and rabbits is unknown, according to the known conditions listed two equations, to work out the number of chickens and rabbits,

 # include & lt; Stdio. H> 
//chicken with rabbit cage, 
Int main (void) 
{
Int a1, b1, c1, a2, b2, c2. 
Int x, y;//the number of chicken rabbit 
Printf (" input the first equation of a1, b1, c1: "); 
The scanf (" % d % d % d ", & amp; A1, & amp; B1, & amp; C1); 
Printf (" \ n input the second equation of a2, b2, c2: "); 
The scanf (" % d % d % d ", & amp; A2, & amp; B2, & amp; C2); 
If b1 b2 - a2 (a1 * *==0)//no judgment formula 
{
Printf (" \ nerror do inpit again. \ n "); 
} 
X=(b2 - c2 c1 * * b1/b2 - a2 (a1 * * b1);//the number of chickens 
Y=c2 - a2 (a1 * * c1)/b2 - a2 (a1 * * b1);//the number of rabbit 
Rabbits chickenes printf (" % d, % d ", x, y); 
return 0; 
} 

The results