PKU 1840 Eqs
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Eqs
Time Limit: 5000MS
Memory Limit: 65536KTotal Submissions: 1818
Accepted: 712
Memory Limit: 65536KTotal Submissions: 1818
Accepted: 712
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.
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
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
a1x13+ a2x23+ a3x33+ a4x43+ a5x53=0
->
a1x13+ a2x23= -a3x33- a4x43- a5x53
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
题目的意思很简单,就是给你几个系数,叫你求解的个数
a1x13+ a2x23+ a3x33+ a4x43+ a5x53=0
->
a1x13+ a2x23= -a3x33- a4x43- a5x53
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
- for(i=1;i<=5;i++)for(j=0;j<=100;j++)A[i][j]=a[i]*tpow[j];
- for(i=-50;i<=50;i++)
- {
- if(!i)continue;
- for(j=-50;j<=50;j++)
- {
- if(!j)continue;
- index=(buf=A[1][i+50]+A[2][j+50]+18750000)%p;
- top.val=buf;top.cnt=1;
- if((tsize=hash[index].size())==0)
- hash[index].push_back(top);
- else
- {
- for(k=0;k<tsize;k++)
- if(buf==hash[index][k].val)
- {
- hash[index][k].cnt++;
- break;
- }
- if(k==tsize)
- hash[index].push_back(top);
- }
- }
- }
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