Equations HDU - 1496（哈希的应用）

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

题目大意：

       给你a,b,c,d这4个数的值，然后问a*x1^2 + b*x2^2 +  c*x3^2 + d*x4^2 = 0

的（x1,x2,x3,x4）解一共有多少种？

思路：

        要想直接暴力4层for循环是不可能的，但可以直接3层暴力；因为正负号不同不影响平方的效果，所以只枚举正数，最后乘以2的4次方（16），表示正负不同的组合：

操作代码如下：

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
using namespace std;
int main()
{
    int a,b,c,d;
    while(~scanf("%d%d%d%d",&a,&b,&c,&d))
    {
        if(a>&&b>&&c>&&d>||a<&&b<&&c<&&d<)
        {
            printf("0\n");
            continue;
        }
        int sum=;
        for(int i=; i<=; i++)
            for(int j=; j<=; j++)
                for(int l=; l<=; l++)
                {
                    int f=a*i*i+b*j*j+c*l*l;
                    if(f%d==)
                    {
                        int h=sqrt(abs((-f)/d));
                        if(h<=&&h>&&h*h==abs((-f)/d)&&f+d*h*h==)
                            sum++;
                    }
                }
        printf("%d\n",sum*);
    }
    return ;
}

另外一种可用哈希进行求解，分别把前两个数所算的总和：正数和负数放入两个不同的数组中，之后与后两个的负数与正数进行匹配看最终有多少种？结果还要  *16 哦！

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define maxx 1001001
int negative[maxx];//记录负数
int positive[maxx];//记录正数
int main()
{
    int a,b,c,d;
    while(~scanf("%d%d%d%d",&a,&b,&c,&d))
    {
        if(a>&&b>&&c>&&d>||a<&&b<&&c<&&d<)
        {//abcd全部大于0或者小于0，肯定无解。要加上这个，不然超时
            printf("0\n");
            continue;
        }
        memset(negative,,sizeof(negative));
        memset(positive,,sizeof(positive));
        for(int i=; i<=; i++)
            for(int j=; j<=; j++)
            {
                int k=a*i*i+b*j*j;
                if(k<=)
                    negative[-k]++;//k<=0  负值[k]++
                else
                    positive[k]++;//k>0 正值++
            }
        int sum=;
        for(int i=; i<=; i++)
            for(int j=; j<=; j++)
            {
                int k=c*i*i+d*j*j;
                if(k<)
                    sum+=positive[-k];//若k为负，加上正值
                else
                    sum+=negative[k]; //若k为正，加上负值
            }
        printf("%d\n",*sum); //每个解有正有负，结果有2^4种
    }
    return ;
}