Equations
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6089 Accepted Submission(s): 2465
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.
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.
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
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<string>
#include<algorithm>
#include<cstdlib>
using namespace std;
int hash1[1000005],hash2[5000005];
int a,b,c,d;
int main()
{
while(scanf("%d%d%d%d",&a,&b,&c,&d)!=EOF)
{
int ans=0,temp;
if((a>0&&b>0&&c>0&&d>0)||(a<0&&b<0&&c<0&&d<0))
{
printf("0
");
continue;
}
for(int i=0;i<1000005;i++)
hash1[i]=hash2[i]=0;
for(int i=1;i<=100;i++)
{
for(int j=1;j<=100;j++)
{
temp=a*i*i+b*j*j;
//printf("%d
",temp);
if(temp>=0)
hash1[temp]++;
else
hash2[-temp]++;
}
}
for(int i=1;i<=100;i++)
{
for(int j=1;j<=100;j++)
{
temp=c*i*i+d*j*j;
if(temp>0)
ans+=hash2[temp];
else
ans+=hash1[-temp];
}
//printf("%d
",ans);
}
printf("%d
",16*ans);
}
return 0;
}