【超级hash大法】HDU 1496 Equations

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


题意：求有多少个解！！

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

#include <iostream>
#include <fstream>
#include <algorithm>
#include <string>
#include <set>
//#include <map>
#include <queue>
#include <utility>
#include <stack>
#include <list>
#include <vector>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <ctime>
#include <ctype.h>
using namespace std;

int hash[3000005];

int main()
{
    int a, b, c, d, x1, x2, x3, x4, res;
    while (~scanf ("%d%d%d%d", &a, &b, &c, &d))
    {
		if (a > 0 && b > 0 && c > 0 && d > 0 || a < 0 && b < 0 && c < 0 && d < 0)    //都是同号必定无解，木有这个超时
		{
			puts ("0");
			continue;
		}
        memset (hash, 0, sizeof(hash));
        res = 0;
        for (x1 = 1; x1 < 101; x1++)    //hash令四重循环变成二重，复杂度大大减少
			for (x2 = 1; x2 < 101; x2++)
				hash[a*x1*x1+b*x2*x2+2000000]++;
		for (x3 = 1; x3 < 101; x3++)
			for (x4 = 1; x4 < 101; x4++)
					res += hash[(0-(c*x3*x3+d*x4*x4))+2000000];
		res *= 16;
   //因为正负号不同不影响x平方的效果，所以直接枚举正数，最后乘以2的4次方，表示正负不同的组合
        printf ("%d\n", res);
    }
    return 0;
}

评论

2 楼 基德KID.1412 2011-08-11  

哈哈！！+2 

1 楼 sgeteternal 2011-08-10  

哈哈！！1