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方程式

2024-07-02 17:11:42   224人阅读

方程式

 

Time Limit:   1000MS       Memory Limit:   65535KB
Submissions:   2312       Accepted:   580

 

Description

Consider equations having the following form: a*x1*x1 + b*x2*x2 + c*x3*x3 + d*x4*x4 = 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.

Output

or 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

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#include<stdio.h>
#include<cstring>
//using namespace std;
#define MAX 1000000
int hash[MAX*2+1];
int main()
{
 int i,j,sum;
 int a,b,c,d,ans;
 while(scanf("%d %d %d %d",&a,&b,&c,&d)!=EOF)
 {
  ////若memset(hash,0,sizeof(hash));放此处会超时
  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));/////////////位置决定是否超时
  for(i=1;i<=100;i++)
  {  
   for(j=1;j<=100;j++)
   {
    sum=a*i*i+b*j*j;
    hash[sum+MAX]++;
   } 
  }
  ans=0;
  for(i=1;i<=100;i++)
  {
    
   for(j=1;j<=100;j++)
   {
    sum=-(c*i*i+d*j*j);
    ans+=hash[sum+MAX];
   }
  }
  ans*=16;
  printf("%d\n",ans);
 }
 return 0;
}

　　


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