# POJ 1804 Eqs

C - Eqs

Time Limit:5000MS     Memory Limit:65536KB     64bit IO Format:%I64d
& %I64u

Submit Status

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

#define maxn 25000005
short int hash[maxn];

int main()
{
int a,b,c,d,e;
while(cin>>a>>b>>c>>d>>e)
{
memset(hash,0,sizeof(hash));

int i1,i2,i3,i4,i5;
int sum=0;
for(i1=-50;i1<=50;i1++)
if(i1)
for(i2=-50;i2<=50;i2++)
if(i2)
{
sum=a*i1*i1*i1+b*i2*i2*i2;
if(sum<0)
sum=sum+maxn;

hash[sum]++;
}

int num=0;

for(i3=-50;i3<=50;i3++)
if(i3)
for(i4=-50;i4<=50;i4++)
if(i4)
for(i5=-50;i5<=50;i5++)
if(i5)
{
sum=c*i3*i3*i3+d*i4*i4*i4+e*i5*i5*i5;
sum=-sum;

if(sum<0)
sum=sum+maxn;

if(hash[sum]!=0)
num=num+hash[sum];
}

cout<<num<<endl;
}
return 0;
}
```

POJ 1804 Eqs

## POJ 1840 Eqs（暴力）

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## poj 1840 Eqs （Hash）

/* 这题10^8的暴力可以出答案 但是 慢..... 有个小小的bug 出题人卡int 却没注意short 用short可以不用hash 直接搞 49428K 313MS */ #include<iostream> #include<cstdio> #include<cstring> #define base 12500000 using namespace std; int a1,a2,a3,a4,a5,x1,x2,x3,x4,x5,ans; short f[125