Equations（哈希）

Link：http://acm.hdu.edu.cn/showproblem.php?pid=1496

Equations

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6091    Accepted Submission(s): 2467

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 -41 1 1 1

 

Sample Output

390880

AC code：

#include<iostream>#include<cmath>#include<climits>#include<cstring>#include<queue>#include<vector>#include<algorithm>#include<string.h>#include<cstdio>#define MAXN 1000010#define LL long longusing namespace std;int ha[MAXN*2];int sum;int main(){    int i,j,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))        {            printf("0\n");        }        else        {            memset(ha,0,sizeof(ha));            for(i=1; i<=100; i++)            {                for(j=1; j<=100; j++)                {                    ha[a*i*i+b*j*j+1000000]++;                }            }            sum=0;            for(i=1; i<=100; i++)            {                for(j=1; j<=100; j++)                {                    sum+=ha[-c*i*i-d*j*j+1000000];                }            }            LL ans=sum*2*2*2*2;            printf("%I64d\n",ans);        }    }    return 0;}

空间优化哈希算法：

AC code：

#include<iostream>#include<cmath>#include<climits>#include<cstring>#include<queue>#include<vector>#include<algorithm>#include<string.h>#include<cstdio>#define MAXN 20010#define LL long longusing namespace std;int f[MAXN],g[MAXN];LL ans;int has(int k){    int t=k%MAXN;    if(t<0)    {        t+=MAXN;    }    while(f[t]!=0&&g[t]!=k)    {        t=(t+1)%MAXN;    }    return t;}int main(){    int a,b,c,d,i,j,s,p;    while(scanf("%d%d%d%d",&a,&b,&c,&d)>0)    {        if(a>0&&b>0&&c>0&&d>0||a<0&&b<0&&c<0&&d<0)        {            printf("0\n");            continue;        }        memset(f,0,sizeof(f));        for(i=1; i<=100; i++)        {            for(j=1; j<=100; j++)            {                s=a*i*i+b*j*j;                p=has(s);                f[p]++;                g[p]=s;            }        }        ans=0;        for(i=1; i<=100; i++)        {            for(j=1; j<=100; j++)            {                s=-c*i*i-d*j*j;                p=has(s);                ans+=f[p];            }        }        printf("%I64d\n",ans*16);    }    return 0;}