E:方程式
题目：
Description
Consider equations having the following form: a*x1*x1 + b*x2*x2 + c*x3*x3 + d*x4*x4 = 0
a, b, c, d are integers from the interval [-50,50] and any of them cannot be 0.
It is consider a solution a system ( x1,x2,x3,x4 ) that verifies the equation, xi is an integer from [-100,100] and xi != 0, any i ∈{1,2,3,4}.
Determine how many solutions satisfy the given equation.
Input
The input consists of several test cases. Each test case consists of a single line containing the 4 coefficients a, b, c, d, separated by one or more blanks.
Output
or each test case, output a single line containing the number of the solutions.
Sample Input
1 2 3 -4 1 1 1 1
Sample Output
39088 0
代码：
#include <stdio.h>
#include <stdlib.h>
int a[10000],b[10000];
int cmp ( const void *a , const void *b )
{
return *(int *)a - *(int *)b;
}
int main()
{
int x[4],i,j,k,count,t;
while(scanf("%d%d%d%d",&x[0],&x[1],&x[2],&x[3])!=EOF)
{
for(k=0,i=1;i<=100;i++)
{
for(j=1;j<=100;j++)
{
a[k]=x[0]*i*i+x[1]*j*j;
b[k]=-1*(x[2]*i*i+x[3]*j*j);
k++;
}
}
qsort(a,10000,sizeof(a[0]),cmp);//快排
qsort(b,10000,sizeof(b[0]),cmp);
count=i=j=0;
while(i<k && j<k)
{
if(a[i]<b[j]) i++;
else if(a[i]>b[j]) j++;
else
{
t=j+1;
count++;
while(t<k && a[i]==b[t]) {count++;t++;}
i++;
}
}
printf("%d\n",count*16);
}
return 0;
}