POJ1840___Eqs（哈希表）

本文出自svitter的blog
——怒草哈希表！

题意

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.
a1~a5，x1~x5分别在[-50~50]取值，x1~x5均不为0,求一共有多少组解。

输入输出分析

写在这里记录一下， 如果动态开辟空间就不用while(~scanf)这种形式了，可能MLE或者回收错误。

Sample Input

37 29 41 43 47
Sample Output

654

算法数据结构分析

把i**3的值求出来打表存，因为肯定要遍历用到无数次。为了简化计算将0点定在坐标为50的地方。

计算出前两组乘积的和存到哈希表中，然后在计算后面三组值取相反数看是否相等。

注意找到数不要立刻break，因为这一个数可以由多组不同的组合求出，所以属于一种答案。

AC代码

//author: svtter
//

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <vector>
#include <map>
#include <algorithm>
#include <queue>
#include <cmath>

#define INF 0xffffff
#define lln long long

#ifdef ONLINE_JUDGE
#define FOI(file) 0
#define FOW(file) 0
#else
#define FOI(file) freopen(file,"r",stdin);
#define FOW(file) freopen(file,"w",stdout);
#endif

using namespace std;

struct hashnode
{
int v;
hashnode *next;
};

int a[5];
int hash[101];

//先前取值3533
#define M 3533
int i, j, k;
int res;
hashnode * Hash[M+5];

void BuildList()
{
for(i = -50; i < 0; i++)
{
hash[i+50] = i*i*i;
hash[50-i] = -hash[i+50];
}
}

void BuildHash()
{
int hashkey;
hashnode * tmp;
for(i = 0; i <= 100; i++)
{
if(i == 50)
continue;
for(j = 0; j <= 100; j++)
{
if(j == 50)
continue;

res = a[0]*hash[i] + a[1]*hash[j];

hashkey = (res > 0 ? res : -res) % M;

tmp = new hashnode;
tmp->v = res;
tmp->next = Hash[hashkey];
Hash[hashkey] = tmp;
}
}
}

int main()
{
//FOI("input");
//FOW("output");
//write your programme here

BuildList();
int ans = 0;
int hashkey;
hashnode *ph;
scanf("%d%d%d%d%d", &a[0], &a[1], &a[2], &a[3], &a[4]);
ans = 0;
BuildHash();
for(i = 0; i <= 100; i++)
{
if(i == 50)
continue;
for(j = 0; j <= 100; j++)
{
if(j == 50)
continue;
for(k = 0; k <= 100; k++)
{
if(k == 50)
continue;
res = a[2]*hash[i] + a[3]*hash[j] + a[4]*hash[k];
res = -res;

hashkey = (res > 0 ? res : -res) % M;
ph = Hash[hashkey];

//
while(ph != NULL)
{
//可能有多组数据
if(ph->v == res)
ans++;
ph = ph->next;
}
}
}
}
printf("%d\n", ans);

return 0;
}