百鸡问题

int countChickens(int n, int g[],int m[],int s[]) //g[],m[],s[]分别存放当钱数为n时的公鸡、母鸡、小鸡的数量
{
int a,b,c,k;
a = b = k = 0;
int n_g = n/5;		//可买到公鸡的最大数量
int n_m = n/3;

for (;a <= n_g;a++){
for (;b <= n_m; b++){
c = n - a - b;
if ((c%3==0)&&(5*a+3*b+c/3==n)){
g[k] = a;
m[k] = b;
s[k] = c;
k++;
}
}
b=0;
}

return k;
}