HDU1496 hash

Equations

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8124    Accepted Submission(s): 3329

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

Author

LL

Source

“2006校园文化活动月”之“校庆杯”大学生程序设计竞赛暨杭州电子科技大学第四届大学生程序设计竞赛

题意：

计算a*x1^2+b*x2^2+c*x3^2+d*x4^2=0有多少解。系数取值[-50,50],不取0，x取值[-100,100],不取0.

代码：

//判断一下当系数的符号都相同时显然无解。可以把四重循环分成两个二重循环，ax1^2+bx2^2=-cx3^2-dx4^2
//只要两边有相同的值时就出现一个解，由于是x^2，每个x对应正负两个解，4个x共有16个解。
//最重要的是hash.
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int a,b,c,d;
int m[];
int main()
{
    while(~scanf("%d%d%d%d",&a,&b,&c,&d)){
        if((a>&&b>&&c>&&d>)||(a<&&b<&&c<&&d<)){
            printf("0\n");
            continue;
        }
        memset(m,,sizeof(m));
        for(int i=;i<=;i++){
            for(int j=;j<=;j++){
                m[a*i*i+b*j*j+]++;
            }
        }
        int ans=;
        for(int i=;i<=;i++){
            for(int j=;j<=;j++){
                ans+=m[-c*i*i-d*j*j+];
            }
        }
        printf("%d\n",ans*);
    }
    return ;
}