POJ 1840 Eqs(hash)

题意  输入a1,a2,a3,a4,a5  求有多少种不同的x1,x2,x3,x4,x5序列使得等式成立   a,x取值在-50到50之间

直接暴力的话肯定会超时的   100的五次方  10e了都    然后可以考虑将等式变一下形   把a1*x1^3+a2*x2^3移到右边   也就是-(a1*x1^3+a2^x2^3)=a3*x3^3+a4*x4^3+a5*x5^3

考虑到a1*x1^3+a2^x2^3的最大值50*50^3+50*50^3=12500000  这个数并不大  可以开这么大的数组把每个结果出现的次数存下来   又因为结果最小可能是负的12500000   负数不能做数组的下标   加上个12500000*2就行了   这样分别枚举左右两边   把左边出现过的结果都存在一个数组里面    再枚举右边   没出现一次结果  答案就加上前面这个结果出现的次数
   枚举完就出现答案了

#include<cstdio>
#include<cstring>
using namespace std;
const int maxs = 50 * 50 * 50 * 50 * 4 + 10;
unsigned short cnt[maxs];
int main()
{
int a1, a2, a3, a4, a5, sum,ans=0;
scanf ("%d%d%d%d%d", &a1, &a2, &a3, &a4, &a5);
for (int x1 = -50; x1 <= 50; ++x1)
{
if (x1 == 0) ++x1;
for (int x2 = -50; x2 <= 50; ++x2)
{
if (x2 == 0) ++x2;
sum = (a1 * x1 * x1 * x1 + a2 * x2 * x2 * x2) * (-1);
if (sum < 0) ++cnt[sum + maxs];
else ++cnt[sum];
}
}

for (int x3 = -50; x3 <= 50; ++x3)
{
if (x3 == 0) ++x3;
for (int x4 = -50; x4 <= 50; ++x4)
{
if (x4 == 0) ++x4;
for (int x5 = -50; x5 <= 50; ++x5)
{
if (x5 == 0) ++x5;
sum = (a3 * x3 * x3 * x3 + a4 * x4 * x4 * x4 + a5 * x5 * x5 * x5) ;
if (sum < 0) sum += maxs;
ans += cnt[sum];
}
}
}
printf ("%d\n", ans);
return 0;
}

Eqs

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

还有用hashmap做的  空间优化了不少

#include<iostream>
#include<cstdio>
#include<cstring>
#include<hash_map>
using namespace std;
int first[50*50*50+10];
int ecnt,w[10005],v[10005],nex[10005];

void add(int x)
{
int t=(x+50*50*50*100)/200,flag=1;
for(int e=first[t];(~e)&&flag;e=nex[e])
{
if(v[e]==x)
{
flag=0,w[e]++;
}
}
if(flag)
{
w[ecnt]=1;
v[ecnt]=x;
nex[ecnt]=first[t];
first[t]=ecnt++;
}
}
int getcnt(int x)
{
if(x>50*50*50*50*2||x<-50*50*50*50*2)return 0;
int t=(x+50*50*50*100)/200;
for(int e=first[t];(~e);e=nex[e])
{
if(v[e]==x)return w[e];
}
return 0;
}
int main()
{
int a1, a2, a3, a4, a5, sum,ans=0;
scanf ("%d%d%d%d%d", &a1, &a2, &a3, &a4, &a5);
memset(first,-1,sizeof first);
ecnt=0;
for (int x1 = -50; x1 <= 50; ++x1)
{
if (x1 == 0) ++x1;
for (int x2 = -50; x2 <= 50; ++x2)
{
if (x2 == 0) ++x2;
sum = (a1 * x1 * x1 * x1 + a2 * x2 * x2 * x2) * (-1);
add(sum);
}
}
for (int x3 = -50; x3 <= 50; ++x3)
{
if (x3 == 0) ++x3;
for (int x4 = -50; x4 <= 50; ++x4)
{
if (x4 == 0) ++x4;
for (int x5 = -50; x5 <= 50; ++x5)
{
if (x5 == 0) ++x5;
sum = (a3 * x3 * x3 * x3 + a4 * x4 * x4 * x4 + a5 * x5 * x5 * x5) ;
ans +=getcnt(sum);
}
}
}
printf ("%d\n", ans);
return 0;
}