Description

Consider equations having the following form: 
a1x13+ a2x23+ a3x33+ a4x43+ a5x53=0 
The coefficients are given integers from the interval [-50,50]. 
It is consider a solution a system (x1, x2, x3, x4, x5) that verifies the equation, xi∈[-50,50], xi != 0, any i∈{1,2,3,4,5}. 

Determine how many solutions satisfy the given equation. 
Input

The only line of input contains the 
 
5 coefficients a1, a2, a3, a4, a5, separated by blanks.
Output

The output will contain on the first line the number of the solutions for the given equation.
Sample Input

37 29 41 43 47
Sample Output

654

　　芒果君：這道題是裸暴力，但是今天我們有一個新的思路，就是用哈希來優化暴力。我們把五項i=1->5 表示為Xi,那麽很明顯X1+X2+X3==-X4-X5，這樣我們把等式左邊存到哈希表，然後讓等式右邊去找左邊，大大減小了枚舉的時間復雜度。

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 #include<map>
 5 #define mod 100003
 6 #define ll long long
 7 using namespace std;
 8 int hl[mod],a1,a2,a3,a4,a5,cnt,ans;
 9 int cal(int x){return x*x*x;}
10 struct H{
11     int val,ne;
12 }Hash[1100000];
13 void
 
 insert(int x)
14 {
15     int key=abs(x)%mod;
16     Hash[++cnt].val=x;
17     Hash[cnt].ne=hl[key];
18     hl[key]=cnt;
19 }
20 int search(int x)
21 {
22     int sum=0;
23     int key=abs(x)%mod;
24     for(int i=hl[key];i;i=Hash[i].ne)    if(Hash[i].val==x) sum++;
25     return sum;
26 }
27 int main()
28 {
29     scanf("%d%d%d%d%d",&a1,&a2,&a3,&a4,&a5);
30     for(int i=-50;i<=50;++i)if(i)
31         for(int j=-50;j<=50;++j)if(j)
32             for(int k=-50;k<=50;++k)if(k)
33                 insert(a1*cal(i)+a2*cal(j)+a3*cal(k));
34     for(int i=-50;i<=50;++i)if(i)
35         for(int j=-50;j<=50;++j)if(j)
36             ans+=search(-a4*cal(i)-a5*cal(j));
37     printf("%d",ans);
38     return 0;
39 }

POJ 1840：Eqs