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

 int main()
 {
 //    freopen("input.txt", "r", stdin);
     int rate[], a, b, c, d, i, j, sum, temp, res;
     for(i=; i<; ++i) rate[i] = i*i;
     while(scanf("%d%d%d%d", &a, &b, &c, &d) != EOF)
     {
         if((a>&&b>&&c>&&d>) || (a<&&b<&&c<&&d<))
         {
             printf("0\n");
             continue;
         }
         memset(num, , sizeof(num));
         for(i=; i<; ++i)
             for(j=; j<; ++j)
             {
                 temp = a*rate[i]+b*rate[j];
                 res = hash(temp);
                 store[res] = temp;
                 num[res]++;
             }

             sum = ;

         for(i=; i<; ++i)
             for(j=; j<; ++j)
             {
                 temp = -(c*rate[i]+d*rate[j]);
                 res = hash(temp);
                 sum += num[res];
             }

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