(HDU - 1496)Equations

(HDU - 1496)Equations

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 8701 Accepted Submission(s): 3615

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

题目大意：给出四个系数a,b,c,d问有多组满足 xi∈[−100,100]且xi≠0且1<=i<=4 的解使得方程 a∗x12+b∗x22+c∗x32+d∗x44=0 成立。

思路：二重循环+hash存储查找。因为对于每一个xi都可以有正有负，所以最后还要乘16.

#include
#include
using namespace std;

const int maxn=2000005;
int hash[maxn];

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));
        for(int i=1;i<=100;i++)
            for(int j=1;j<=100;j++)
                hash[a*i*i+b*j*j+1000000]++;
        int ans=0;
        for(int i=1;i<=100;i++)
            for(int j=1;j<=100;j++)
                ans+=hash[-c*i*i-d*j*j+1000000];
        printf("%d\n",ans*16);
    }
    return 0;
}