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！！！
代码：
Cpp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;
}
补充：软件开发 , C语言 ,
