ACM题目————Equations

Description

Consider equations having the following form:

a*x1^2+b*x2^2+c*x3^2+d*x4^2=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.
End of file.

 

Output

For 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

 

哈希只会一丢丢。就不写详解了。自己体会，O(∩_∩)O哈哈~

 

对于最后的结果要乘16，是因为，计算出来的结果可正可负，一共有 2^4 种情况

 

?

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

using namespace std; const int maxn = 2000005; const int h = 1000000; int hash1[maxn]; int a, b, c, d, sum, k;   //主思路： 分两段计算 最后乘16（ 2^4 ） int main() {     while(~scanf("%d %d %d %d",&a,&b, &c, &d) )     {         if( a>0 && b>0 && c>0 && d>0 ||             a<0 && b<0 && c<0 && d<0 )         {             cout << "0" << endl ;             continue ;         }         memset(hash1,0,sizeof(hash1));         sum = 0 ;         for(int i=1; i<=100; i++)             for(int j=1; j<=100; j++)                 hash1[ a * i * i + b * j * j + h ] ++ ;         for(int i=1; i<=100; i++)             for(int j=1; j<=100; j++)                 sum += hash1[ h - c * i * i - d * j * j ] ;         cout << sum*16 << endl ;     }       return 0; }