hdu 1496 -- Equations
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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 -41 1 1 1
Sample Output
390880
#include <iostream>using namespace std;int base[101];int hash[2000009];int main(){int a,b,c,d;int i,j,k;for(i=0;i<101;i++)base[i]=i*i;while(scanf("%d %d %d %d",&a,&b,&c,&d)!=EOF){if( (a>0 && b>0 && c>0 && d>0 )|| (a<0 && b<0 && c<0 && d<0) ){ printf("0/n"); continue; }else{memset(hash,0,sizeof(hash));for(i=1;i<101;i++)for(j=1;j<101;j++)hash[base[i]*a+base[j]*b+1000000]++;int sum=0;for(i=1;i<101;i++)for(j=1;j<101;j++)sum+=hash[-(c*base[i]+d*base[j])+1000000];cout<<sum*16<<endl;//X分正负,每个X有2个解,4个就有16种不同的解}}}
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