POJ-1840 Eqs
| Time Limit: 5000MS | Memory Limit: 65536K | |
| Total Submissions: 14929 | Accepted: 7329 |
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.
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
分析：很容易想到半路相遇法，枚举前两个变量的值并放入hash表中，然后再枚举后三个值查询并累加结果。
分析：很容易想到半路相遇法，枚举前两个变量的值并放入hash表中，然后再枚举后三个值查询并累加结果。
#include <cstdio>
#include <cmath>
#include <iostream>
using namespace std;
int a1,a2,a3,a4,a5,ans,f[2000007],val[2000007];
bool jud[2000007];
void insert(int x)
{
int k=abs(x) % 1300707;
while(jud[k] && f[k] != x) k++;
if(!jud[k])
{
jud[k]=true;
f[k]=x;
}
val[k]++;
}
int Find(int x)
{
int k=abs(x) % 1300707;
while(jud[k] && f[k] != x) k++;
return(val[k]);
}
int main()
{
cin.sync_with_stdio(false);
cin>>a1>>a2>>a3>>a4>>a5;
for(int i=-50;i <= 50;i++)
for(int j=-50;j <= 50;j++)
if(i*j) insert(-(a4*i*i*i+a5*j*j*j));
for(int i=-50;i <= 50;i++)
for(int j=-50;j <= 50;j++)
for(int k=-50;k <= 50;k++)
if(i*j*k && abs(a1*i*i*i+a2*j*j*j+a3*k*k*k) <= 2*6250000) ans+=Find(a1*i*i*i+a2*j*j*j+a3*k*k*k);
cout<<ans<<endl;
}