poj 1840 Eqs

http://poj.org/problem?id=1840

题意：a1x1³+ a2x2³+ a3x3³+ a4x4³+ a5x5³=0 ，给出a1,a2,a3,a4,a5,且xi在-50到50（除零外），有多少种满足条件的组合；

思路：用hash,这里有两种，一种用map建立，不过我刚开始超时了，后来发现   for (i = -50; i <= 50; i++)   x[m++] = i * i * i; 先这样处理下就行了，不过时间还是挺长的，另一种就是普通的hash了，这里我用的静态加挂链的方式；

代码：

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
const int qq = 1000100;
struct node
{
    int data;
    int num;
    bool biao;
    node *next;
}link[qq];
int n = 0;
int a[6] = {0};
int temp;
node *get()
{
    node *tt;
    tt = new node;
    tt->next = NULL;
    tt->num = 1;
    return tt;
}
void ff(int temp,node *t,int biao)
{
    if(t->data == temp)
    t->num++;
    else
    if(t->next != NULL)
    ff(temp,t->next,biao);
    else
    {
        node *p;
        p=get();
        p->data = temp;
        p->biao = biao;
        t->next = p;
    }
}
int fin(int temp,node *p,int biao)
{
    if(p->data == temp && p->biao == biao)
    return p->num;
    else
    if(p->next != NULL)
    return fin(temp,p->next,biao);
    else
    return 0;
}
int main()
{
    memset(link,0,sizeof(link));
    for(int i = 1;i <= 5; ++i)
    scanf("%d",&a[i]);
    for(int i = -50; i <= 50; ++i)
        if(i != 0)
        {
            for(int j = -50; j <= 50; ++j)
            if(j != 0)
            {
                for(int k = -50; k <= 50; ++k)
                if(k != 0)
                {
                    int biao = 0;
                    temp = a[1]*i*i*i+a[2]*j*j*j+a[3]*k*k*k;
                    if(temp < 0)
                    {
                        biao = 1;
                        temp = -temp;
                    }
                    if(link[temp%qq].num)
                    ff(temp,&link[temp%qq],biao);
                    else
                    {
                        link[temp%qq].data = temp;
                        link[temp%qq].next = NULL;
                        link[temp%qq].num = 1;
                        link[temp%qq].biao = biao;
                    }
                }
            }
        }
        __int64 ans = 0;
        for(int i = -50; i <= 50; ++i)
        if(i != 0)
        {
            for(int j = -50; j <= 50; ++j)
            {
                if(j == 0)
                continue;
                int biao = 1;
                temp = a[4]*i*i*i+a[5]*j*j*j;
                if(temp <= 0)
                {
                    temp = -temp;
                    biao = 0;
                }
                if(link[temp%qq].num)
                ans = ans + fin(temp,&link[temp%qq],biao);
            }
        }
        printf("%I64d\n",ans);
        return 0;
}