HDU 1496 Equations

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

 

题解：整数哈希，最后乘以16（每个X的值都有正负两个值，组合起来就是2^4个）。

 

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;

const int maxn=50005;
int hashtb[maxn],sum[maxn],tsm[105];

int Hash(int num){
    int temp=num%maxn;
    if(temp<0) temp+=maxn;
    while(sum[temp]!=0&&hashtb[temp]!=num)
        temp=(temp+1)%maxn;
    return temp;
}

int main()
{
    int a,b,c,d;
    for(int i=1;i<=100;i++)
        tsm[i]=i*i;
    while(scanf("%d %d %d %d",&a,&b,&c,&d)!=EOF){
        if(a>0&&b>0&&c>0&&d>0||a<0&&b<0&&c<0&&d<0){
            printf("0\n");
            continue;
        }
        memset(sum,0,sizeof(sum));
        for(int i=1;i<=100;i++){
            for(int j=1;j<=100;j++){
                int ans=a*tsm[i]+b*tsm[j];
                int p=Hash(ans);
                hashtb[p]=ans;
                sum[p]++;
            }
        }
        int tot=0;
        for(int i=1;i<=100;i++){
            for(int j=1;j<=100;j++){
                int ans=-(c*tsm[i]+d*tsm[j]);
                int p=Hash(ans);
                tot+=sum[p];
            }
        }
        printf("%d\n",tot*16);
    }
    return 0;
}