HDOJ Equations

 

Equations

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

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1886    Accepted Submission(s): 717

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

 

//先求出a*x1*x1+b*x2*x2所有的值保存在hash表中，然後查找-(c*x3*x3+d*x4*x4)是否存在hash表中
//注意輸出時需要乘上2^4  
#include<iostream>
#include<cstring>
using namespace std;
int hash[2000050];
int main(){
    int a,b,c,d,x1,x2,x3,x4,num;
    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(hash,0,sizeof(hash));
           for(int i=1;i<=100;i++)
                 for(int j=1;j<=100;j++)
                       hash[1000000+a*i*i+b*j*j]++;      
           num=0;
           for(int i=1;i<=100;i++)
                 for(int j=1;j<=100;j++)
                      num+=hash[1000000+(-c*i*i-d*j*j)];
           printf("%d\n",num*16);           
    }
 return 0;
}

上面的代碼有點浪費空間，百度了一個代碼，挺好的，如下：

#include<iostream>
using namespace std;
const int MAXN=20021; //ax1^2+bx2^2 和 cx3^2+dx4^2分別最多有10000種可能
int hash[MAXN],counter[MAXN];
//確定在哈希表中的位置，處理衝突
int hash_find(int num){
    int t;
    t=num%MAXN;
    if(t<0)
        t+=MAXN;
    while(counter[t]&&hash[t]!=num) //當哈希表內已保存數並且該數不是num  
          t=(t+1)%MAXN; 
    return t;  
}
int main(){
    int a,b,c,d,i,j,ans,t;
    int num,square[101];
    for(i=1;i<=100;i++)
         square[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(counter,0,sizeof(counter));
            for(i=1;i<=100;i++)
                  for(j=1;j<=100;j++){
                         num=a*square[i]+b*square[j];
                         t=hash_find(num);
                         hash[t]=num;
                         counter[t]++;                   
                  }
            ans=0;
            for(i=1;i<=100;i++)
                  for(j=1;j<=100;j++){
                         num=-(c*square[i]+d*square[j]);
                         t=hash_find(num);
                         ans+=counter[t];                  
                  }  
            ans*=16;  //x1,x2,x3,x4不同的正負 
            printf("%d\n",ans);                          
   }
   return 0;
}