hdu 1496,枚举
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
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<vector>
#include<cstdlib>
#include<algorithm>
using namespace std;
#define LL long long
#define ULL unsigned long long
#define UINT unsigned int
#define MAX_INT 0x7fffffff
#define MAX_LL 0x7fffffffffffffff
#define MAX(X,Y) ((X) > (Y) ? (X) : (Y))
#define MIN(X,Y) ((X) < (Y) ? (X) : (Y))
int main(){
int a,b,c,d;
while(scanf(" %d %d %d %d",&a,&b,&c,&d)==4){
if((a<0 && b<0 && c<0 && d<0) ||
(a>0 && b>0 && c>0 && d>0)){
printf("0\n");
continue;
}
int i,j,k;
LL ans=0;
for(i=1; i<=100; i++)
for(j=1; j<=100; j++)
for(k=1; k<=100; k++){
int tmp=b*i*i + c*j*j + d*k*k;
if(MIN(tmp,a) <0 && MAX(tmp,a)>0){
int post=MAX(tmp,-tmp), posa=MAX(a,-a);
if(post%posa==0){
int x=(int)sqrt(0.5+post/posa);
if(x*x == post/posa && x<=100)
ans+=16;
}
}
}
printf("%lld\n",ans);
}
return 0;
}
