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[1000005];
int s2[1000005];
int main(){
    
    while(scanf("%d%d%d%d",&a,&b,&c,&d)!=EOF)
    {
        res=0;
        if((a>0&&b>0&&c>0&&d>0)||((a<0&&b<0&&c<0&&d<0)))
        {
            printf("0
");
            continue;
        }
        memset(s1,0,sizeof(s1));
        memset(s2,0,sizeof(s2));
        for(int i=1;i<=100;i++)
            for(int j=1;j<=100;j++)
            {
                int k=a*i*i+b*j*j;
                if(k>=0)    s1[k]++;
                else    s2[-k]++;
            }
            
        for(int i=1;i<=100;i++)
            for(int j=1;j<=100;j++)
            {
                int k=c*i*i+d*j*j;
                if(k>0)    res+=s2[k];
                else    res+=s1[-k];                
            }
        
        printf("%d
",res*16);
    }
    
    return 0;
        
}

 折半枚举

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const int MAXN=10005;
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>0&&b>0&&c>0&&d>0||a<0&&b<0&&c<0&&d<0)
        {
            printf("0
");
            continue;
        }
        int res=0;
        int cnt=0;
        for(int i=1;i<=100;i++)
            for(int j=1;j<=100;j++)
                num[cnt++]=a*i*i+b*j*j;
        sort(num,num+cnt);
        for(int i=1;i<=100;i++)
            for(int j=1;j<=100;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
",res*16);
    }
    
    return 0;
}