HDU 1496 Equations

Equations

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2879    Accepted Submission(s): 1114

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

 

 

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

 

 

Recommend

LL

//有时感觉自己就是个水货、自己坑自己

//算是hash的简单应用吧

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
int h[103];
bool f[10003];
int main()
{
    int a,b,c,d;
    int i,j,k,temp,t;
    for(i=1;i<=100;i++)
      h[i]=i*i,f[i*i]=1;
    while(scanf("%d%d%d%d",&a,&b,&c,&d)!=EOF)
    {
        if(a>0&&b>0&&c>0&&d>0) { printf("0\n");continue;}
        if(a<0&&b<0&&c<0&&d<0) { printf("0\n");continue;}
        if(a==0||b==0||c==0||d==0) { printf("0\n");continue;}
        int n=0;
        for(i=1;i<=100;i++)
          for(j=1;j<=100;j++)
            for(k=1;k<=100;k++)
              {
                   temp=a*h[i]+b*h[j]+c*h[k];
                   t=d;
                   if((temp>0&&t>0)||(temp<0&&t<0)) continue;
                   if(temp==0) continue;
                   if(temp<0) temp=-temp;
                   if(t<0) t=-t;
                   if(temp%t!=0) continue;
                   temp=temp/t;
                   if(temp>10000||!f[temp]) continue;
                   n++;
              }
        printf("%d\n",n*16);//自己很2的写个“n=”在输出里、还一直不明白为啥WA了、看到有这想抽自己的冲动

     //n*16 原因是现在求出来的都是正解，每个x都有正负2种情况 就是2^4=16咯，自己都差点忘记了
    }
    return 0;
}