G - Equations

Time Limit: 3000 MS Memory Limit: 32768 KB

64-bit integer IO format: %I64d , %I64u Java class name: Main

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

题意：

给定a,b,c,d让求满足a*x1*x1+b*x2*x2+c*x3*x3+d*x4*x4=0的解的个数 xi != 0。

思路：

n^4暴力肯定超时，可以把方程分成两个部分，t=a*x1*x1+b*x2*x2所有可能的答案遍历一遍，如果t>=0，就将zh[t]+1，否则将fu[-t]+1，再把t=c*x3*x3+d*x4*x4的值遍历一遍，若t>0,就看fu[t]中的解的个数，否则看zh[-t]。最后将结果乘16，因为每一个x都有一个正数一个负数可取，即2^4个。

代码：

#include <cstdio>
#include <cstring>
#include <iostream>
#define LL long long
using namespace std;
int fu[1000005], zh[1000005];
int main()
{
    int a, b, c, d;
    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))    ///若a,b,c,d同号，则一定无解
            printf("0\n");
        else{
            memset(fu, 0, sizeof(fu));
            memset(zh, 0, sizeof(zh));
            for(int i=1; i<=100; i++){
                for(int j=1; j<=100; j++){
                    int temp = a*i*i + b*j*j;
                    if(temp >= 0) zh[temp] ++;    ///记录正
                    else fu[-temp] ++;    ///记录负
                }
            }
            int ans = 0;
            for(int i=1; i<=100; i++){
                for(int j=1; j<=100; j++){
                    int temp = c*i*i + d*j*j;
                    if(temp > 0) ans += fu[temp];    ///查找正
                    else ans += zh[-temp];    ///查找负
                }
            }
            printf("%d\n", ans*16);    ///结果有2^4=16个
        }
    }
}

HDU 1496 (Hash)