Equations

Equations

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

Total Submission(s): 82 Accepted Submission(s): 54

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

 

Sample Output

39088
0

 

Author

LL

 

Source

“2006校园文化活动月”之“校庆杯”大学生程序设计竞赛暨杭州电子科技大学第四届大学生程序设计竞赛

 

Recommend

LL

 

/*
题意：给你a,b,c,d然后让你输出满足条件的 x1,x2,x3,x4 的组数

初步思路：打表乱搞一下

#放弃了：打了半小时了，还没有打完。想一下可以将整个式子分成两半，求两半加起来是零的个数
*/
#include<bits/stdc++.h>
using namespace std;
int a,b,c,d;
int f1[1000005];//正的结果
int f2[1000005];//负的结果
int main(){
    // freopen("in.txt","r",stdin);
    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
");
            continue;
        }
        memset(f1,0,sizeof f1);
        memset(f2,0,sizeof f2);
        long long s=0;
        for(int i=-100;i<=100;i++){
            if(i==0) continue;
            for(int j=-100;j<=100;j++){
                if(j==0) continue;
                int k=a*i*i+b*j*j;
                if(k>=0) f1[k]++;
                else f2[-k]++;
            }
        }
        for(int i=-100;i<=100;i++){
            if(i==0) continue;
            for(int j=-100;j<=100;j++){
                if(j==0) continue;
                int k=c*i*i+d*j*j;
                if(k>0) s+=f2[k];
                else s+=f1[-k];
            }
        }
        printf("%lld
",s);
    }
    return 0;
}