hdu 1496 Equations

Equations

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 6262 Accepted Submission(s): 2536



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 -4
1 1 1 1这题是要用hash法，第一次学hash， 参考了大神代码，
2014,6,24
#include<stdio.h>
#include<string.h>
int x[1000005];//保存得数是正的
int y[1000005];//保存得数是负的
int main(){
int i,j,k,l,count,a,b,c,d;
while(~scanf("%d%d%d%d",&a,&b,&c,&d)){
count=0;
if((a>0&&b>0&&c>0&&d>0)||(a<0&&b<0&&c<0&&d<0)){
printf("0\n");
continue;
}
memset(x,0,sizeof(x));
memset(y,0,sizeof(y));
for(i=1;i<=100;i++)
for(j=1;j<=100;j++){
k=a*i*i+b*j*j;
if(k>=0) x[k]++;//这里k要>=0
else y[-k]++;
}
for(i=1;i<=100;i++)
for(j=1;j<=100;j++){
k=c*i*i+d*j*j;
if(k>0) count+=y[k];//若k为正，加上的f2[k]  ，这里k要>0，不能>=0
else count+=x[-k];//若k为负，加上的f1[k]
}
printf("%d\n",count*16);//每个解有正有负，结果有2^4种
}
return 0;
}

Sample Output

39088
0