HDU ACM 1496 Equations

Equations

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3978    Accepted Submission(s): 1602

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

 

Author

LL

 

Recommend

LL

 

解题思路：暴力+hash，主要是想练练hash，一直认为hash是高端黑，用线性探测再散列处理冲突，记得不知道谁说过处理余数的数一般用上素数，不要忘了最后乘上16，因为四个数有可能是正负

#include<stdio.h>
#include<string.h>
#define MAXN 50001
int num[MAXN], store[MAXN];

int hash(int cur)
{
    int temp = cur%MAXN;
    if(temp < 0) temp += MAXN;
    while(num[temp] != 0 && store[temp] != cur)
    {
        temp = (temp+1)%MAXN;
    }
    return temp;
}

int main()
{
//    freopen("input.txt", "r", stdin);
    int rate[101], a, b, c, d, i, j, sum, temp, res;
    for(i=0; i<101; ++i) rate[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
");
            continue;
        }
        memset(num, 0, sizeof(num));
        for(i=1; i<101; ++i)
            for(j=1; j<101; ++j)
            {
                temp = a*rate[i]+b*rate[j];
                res = hash(temp);
                store[res] = temp;
                num[res]++;
            }
            
            sum = 0;
            
        for(i=1; i<101; ++i)
            for(j=1; j<101; ++j)
            {
                temp = -(c*rate[i]+d*rate[j]);
                res = hash(temp);
                sum += num[res];
            }
            
        printf("%d
", sum*2*2*2*2);
    }
    return 0;
}