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
題目大意，給你a,b,c,d這4個數的值，然後問a*x1^2 + b*x2^2 + c*x3^2 + d*x4^2 = 0
的（x1,x2,x3,x4）解一共有多少種？
初看這題，想直接4次循環找，但是這樣絕對超時，所以就用了hash這種方法來解決，很巧妙！分開兩部分求和，若兩部分的和是0，則就加上那麼多種，最後乘以16。這樣就能從n^4變成2*n^2，速度快了很多很多！！！
代碼：
Cpp代碼
#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <memory.h>
using namespace std;
int f1[1000005]; //保存得數是正的
int f2[1000005]; //保存得數是負的
int main()
{
int i, j, k, sum;
int a, b, c, d;
while(scanf("%d %d %d %d", &a, &b, &c, &d) != EOF)
{
//abcd全部大於0或者小於0，肯定無解。要加上這個，不然超時
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為正，加上的f2[k]
else sum += f1[-k]; //若k為負，加上的f1[k]
}
}
printf("%d\n", 16*sum); //每個解有正有負，結果有2^4種
}
return 0;
}
