poj 1840 Eqs(Hash)

Eqs

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 13829		Accepted: 6787

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

题意：

x1,x2,x3,x4,x5未知，求使方程式成立的情况有多少种。

思路：

看题目就知道数据有点大，如果5个for循环的话就是100^5，1ms处理1000条信息，绝对超时。转换一下方程式，即-（a1x13+
a2x23）=a3x33+
a4x43+ a5x53 。两个for循环，小于零标记并计数。在3层for循环，记录使等式成立的结果。

代码：

#include<iostream>
#include<cstring>
using namespace std;

short hash[25000001];

int main(void)
{
int a1,a2,a3,a4,a5;
while(cin>>a1>>a2>>a3>>a4>>a5)
{
memset(hash,0,sizeof(hash));

for(int x1=-50;x1<=50;x1++)
{
if(!x1)
continue;

for(int x2=-50;x2<=50;x2++)
{
if(!x2)
continue;
int sum=(a1*x1*x1*x1 + a2*x2*x2*x2)*(-1);
if(sum<0)
sum+=25000000;
hash[sum]++;
}
}

int num=0;

for(int x3=-50;x3<=50;x3++)
{
if(!x3)
continue;
for(int x4=-50;x4<=50;x4++)
{
if(!x4)
continue;
for(int x5=-50;x5<=50;x5++)
{
if(!x5)
continue;
int sum=a3*x3*x3*x3 + a4*x4*x4*x4 + a5*x5*x5*x5;
if(sum<0)
sum+=25000000;
if(hash[sum])
num+=hash[sum];
}
}
}
cout<<num<<endl;
}
return 0;
}