9.3.8 Equations

Equations

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 50 Accepted Submission(s): 36

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

思路：Hash

分兩部分Hash，完了

最後結果*16 由於每一個數正負均可以，2^4=16

 1 /*
 2 Author:wuhuajun
 3 */
 4 #include <cmath>
 5 #include <cstdio>
 6 #include <algorithm>
 7 #include <cstring>
 8 #include <string>
 9 #include <cstdlib>
10 using namespace std;
11 
12 typedef long long ll;
13 typedef double dd;
14 const int maxn=2000010;
15 int ff[maxn];
16 int *f=ff+1000003;
17 int a,b,c,d,sum,cnt,ans;
18 
19 void close()
20 {
21 exit(0);
22 }
23 
24 
25 void init()
26 {
27     while (scanf("%d %d %d %d",&a,&b,&c,&d)!=EOF)
28     {
29         if (a>0 && b>0 && c>0 && d>0)
30         {
31             printf("0\n");
32             continue;
33         }
34         if (a<0 && b<0 && c<0 && d<0)
35         {
36             printf("0\n");
37             continue;
38         }
39         memset(ff,0,sizeof(ff));
40         for (int x1=1;x1<=100;x1++)
41         {
42             if (x1==0) continue;
43             for (int x2=1;x2<=100;x2++)
44             {
45                 if (x2==0) continue;
46                 //printf("%d\n",a*x1*x1+b*x2*x2);
47                 f[a*x1*x1+b*x2*x2]++;
48             }
49         }
50         ans=0;
51         for (int x3=1;x3<=100;x3++)
52         {
53             if (x3==0) continue;
54             for (int x4=1;x4<=100;x4++)
55             {
56                 if (x4==0) continue;
57                 sum=c*x3*x3+d*x4*x4;
58                 //printf("x3:%d x4:%d sum:%d\n",x3,x4,sum);
59                 ans+=f[-sum]*16;
60             }
61         }
62         printf("%d\n",ans);
63     }
64 }
65 
66 int main ()
67 {
68     init();
69     close();
70     return 0;
71 }