http://acm.hdu.edu.cn/showproblem.php?pid=1496

思路：主要將等式化為左右兩部分，用一個hash數組先把左邊的值存起來，，然後在計算右面的值時只需要定址就行了，，，，

Equations

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2456 Accepted Submission(s): 969

Problem DescriptionConsider 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.

InputThe 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.

OutputFor each test case, output a single line containing the number of the solutions.

Sample Input

1 2 3 -41 1 1 1

Sample Output

390880

AuthorLL

AC代碼：

#include<iostream>#include<string.h>#define N 1000000using namespace std;char Hash[2*N+100];int main(){ int a,b,c,d;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))//剪枝，，，，，，，{printf("0\n");continue;}memset(Hash,0,sizeof(Hash));for(int i=1;i<=100;++i)for(int j=1;j<=100;++j)Hash[1000000+a*i*i+b*j*j]++;     int sum=0;for(int i=1;i<=100;++i)for(int j=1;j<=100;++j)sum+=Hash[1000000-c*i*i-d*j*j];printf("%d\n",16*sum);}return 0;}