HDU1496(巧妙hash)

Equations

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7057    Accepted Submission(s): 2858

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

以空间换时间。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int a,b,c,d;
int res;
int s1[];
int s2[];
int main(){

    while(scanf("%d%d%d%d",&a,&b,&c,&d)!=EOF)
    {
        res=;
        if((a>&&b>&&c>&&d>)||((a<&&b<&&c<&&d<)))
        {
            printf("0\n");
            continue;
        }
        memset(s1,,sizeof(s1));
        memset(s2,,sizeof(s2));
        for(int i=;i<=;i++)
            for(int j=;j<=;j++)
            {
                int k=a*i*i+b*j*j;
                if(k>=)    s1[k]++;
                else    s2[-k]++;
            }

        for(int i=;i<=;i++)
            for(int j=;j<=;j++)
            {
                int k=c*i*i+d*j*j;
                if(k>)    res+=s2[k];
                else    res+=s1[-k];
            }

        printf("%d\n",res*);
    }

    return ;

}

折半枚举

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const int MAXN=;
int a,b,c,d;
int res;
int num[MAXN];
int main(){

    while(scanf("%d%d%d%d",&a,&b,&c,&d)!=EOF)
    {
        if(a>&&b>&&c>&&d>||a<&&b<&&c<&&d<)
        {
            printf("0\n");
            continue;
        }
        int res=;
        int cnt=;
        for(int i=;i<=;i++)
            for(int j=;j<=;j++)
                num[cnt++]=a*i*i+b*j*j;
        sort(num,num+cnt);
        for(int i=;i<=;i++)
            for(int j=;j<=;j++)
                res+=(upper_bound(num,num+cnt,-(c*i*i+d*j*j))-lower_bound(num,num+cnt,-(c*i*i+d*j*j)));
        printf("%d\n",res*);
    }

    return ;
}