HDU 1496 hash

Equations

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6800    Accepted Submission(s): 2760

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 -41 1 1 1

 

为0的时候注意....

Sample Output

390880
 
#include<cstdio>#include<iostream>#include<cstring>using namespace std;int f1[1000001]={0};int f2[1000001]={0};int main(){ int a=0,b=0,c=0,d=0; while(scanf("%d%d%d%d",&a,&b,&c,&d)!=EOF) {  memset(f1,0,sizeof(f1));  memset(f2,0,sizeof(f2));  if(a>0&&b>0&&c>0&&d>0||a<0&&b<0&&c<0&&d<0)  {   printf("0\n");   continue;  }    for(int i=1;i<=100;i++)   {    for(int j=1;j<=100;j++)    {     int k=a*i*i+b*j*j;     if(k>=0)     {      f1[k]++;   }   else   {    f2[-k]++;   }  }   }   int sum=0;  for(int i=1;i<=100;i++)   {    for(int j=1;j<=100;j++)    {     int k=c*i*i+d*j*j;     if(k>0)     {      sum+=f2[k];   }   else   {    sum+=f1[-k];   }  }   }   printf("%d\n",16*sum); } return 0;}