（hash）hdu 1496 Equations

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 on e 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

解题报告：

题目意思很好理解，就是求有多少种方程的解的形式，输出来就可以了。

如果用4重循环的话，可能会超时，没有试过。。直接用hash一一对应就可以了。

2重循环把s = a*x1*x1+b*x2*x2 看做一个数组的下标，如果s > 0 ,hash1[s]++;若s <= 0; 则 hash2[-s]++;

再2重循环把s = c*x3*x3+d*x4*x4 看做下标，开始查找。多的不说直接代码。。很容易懂的、。、

co de：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int hash1[1000002], hash2[1000002];
int main()
{
    int a, b, c, d, i, j;    
    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;
      }
        int ans = 0, s;
        for(i = 0; i <= 1000001; i++)
            hash1[i] = hash2[i] = 0;
        for(i = 1; i <= 100; i++)
        {
            for(j = 1; j <= 100; j++)
            {
                s = a * i * i + b * j * j;
                if(s > 0)
                    hash1[s]++;
                else
                    hash2[-s]++;
            }
        }
        for(i = 1; i <= 100; i++)
        {        
            for(j = 1; j <= 100; j++)
            {
                s = c * i * i + d * j * j;
                if(s >= 0)
                 ans += hash2[s];
                else
                 ans += hash1[-s];
            }
        }
        printf("%d\n", 16 * ans);
    }
return 0;
}