Equations HDU - 1496 （哈希）

里云“智慧航空AI大赛”（报名中...）

Equations Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 8648    Accepted Submission(s): 3583 Problem 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   Author LL

题目分析：枚举前两个放入哈希数组中，然后判断另外两个的相反数在不在数组中，如果在则有一组符合。

代码：

#include 
#include 
 
int hash[2000005];
 
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 )
        {
            printf( "0\n" );
            continue;
        }
        memset( hash, 0, sizeof( hash ) );
        int cnt= 0;
        for( int i= 1; i<= 100; ++i )
        {
            for( int j= 1; j<= 100; ++j )
            {
                hash[ a* i* i+ b* j* j+ 1000000 ]++;
            }
        }
        for( int i= 1; i<= 100; ++i )
        {
            for( int j= 1; j<= 100; ++j )
            {
                cnt+= hash[ 1000000- c*i *i- d*j *j ];
            }
        }
        printf( "%d\n", cnt* 16 );
    }
     
}