hdu 1496 Equations（非常巧妙的hash）

Equations

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1729    Accepted Submission(s): 662

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？

        初看这题，想直接4次循环找，但是这样绝对超时，所以就用了hash这种方法来解决，很巧妙！分开两部分求和，若两部分的和是0，则就加上那么多种，最后乘以16。这样就能从n^4变成2*n^2，速度快了很多很多！！！

链接：http://acm.hdu.edu.cn/showproblem.php?pid=1496

代码：

#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <memory.h>
using namespace std;

int f1[1000005];    //保存得数是正的
int f2[1000005];    //保存得数是负的

int main()
{
    int i, j, k, sum;
    int a, b, c, d;
    while(scanf("%d %d %d %d", &a, &b, &c, &d) != EOF)
    {
        //abcd全部大于0或者小于0，肯定无解。要加上这个，不然超时
        if(a>0 && b>0 && c>0 && d>0 || a<0 && b<0 && c<0 && d<0)
        {
            printf("0\n");
            continue;
        }
        memset(f1, 0, sizeof(f1));
        memset(f2, 0, sizeof(f2));
        for(i = 1; i <= 100; i++)
        {
            for(j = 1; j<= 100; j++)
            {
                k = a*i*i + b*j*j;
                if(k >= 0) f1[k]++; //k>=0 f1[k]++
                else f2[-k]++;      //k<0  f2[k]++
            }
        }
        sum = 0;
        for(i = 1; i <= 100; i++)
        {
            for(j = 1; j<= 100; j++)
            {
                k = c*i*i + d*j*j;
                if(k > 0) sum += f2[k]; //若k为正，加上的f2[k]
                else sum += f1[-k];     //若k为负，加上的f1[k]
            }
        }
        printf("%d\n", 16*sum); //每个解有正有负，结果有2^4种
    }

    return 0;
}