PKU 1840 Eqs

Eqs

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 1818		Accepted: 712

Description

Consider equations having the following form:
a1x1 ³+ a2x2 ³+ a3x3 ³+ a4x4 ³+ a5x5 ³=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

Source

Romania OI 2002

题目的意思很简单，就是给你几个系数，叫你求解的个数
a1x1 ³+ a2x2 ³+ a3x3 ³+ a4x4 ³+ a5x5 ³=0
->
a1x1 ³+ a2x2 ³= -a3x3 ³- a4x4 ³- a5x5 ³
因此枚举2边.
其中重点是HASH
在ZX大牛的指导下偶学会了如何解决冲突
(我用的HASH很土- -!!!)

核心代码如下

1. for(i=1;i<=5;i++)for(j=0;j<=100;j++)A[i][j]=a[i]*tpow[j];
2.     for(i=-50;i<=50;i++)
3.     {
4.         if(!i)continue;
5.         for(j=-50;j<=50;j++)
6.         {
7.             if(!j)continue;
8.             index=(buf=A[1][i+50]+A[2][j+50]+18750000)%p;
9.             top.val=buf;top.cnt=1;
10.             if((tsize=hash[index].size())==0)
11.                 hash[index].push_back(top);
12.             else
13.             {
14.                 for(k=0;k<tsize;k++)
15.                     if(buf==hash[index][k].val)
16.                     {
17.                         hash[index][k].cnt++;
18.                         break;
19.                     }
20.                 if(k==tsize)
21.                     hash[index].push_back(top);
22.             }
23.         }
24.     }