POJ_1840Eqs解题报告

Eqs

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 7687		Accepted: 3750

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

题目连接：http://poj.org/problem?id=1840

算法类型：hash&&枚举

解题思路：先枚举前三个数的所有值，利用hash散列表分散到一个大数组里，在枚举后2个数的所有值，在和前3个数的所有值相对应，找出所有解。

算法实现：

#include<stdio.h>
#include<string.h>
#define MAX 37500001   //一个很大的质数，大于5个数解的最大值；
char hash[MAX];
int a[6];
void search()
{
int i,j,k,w,q,s,sum=0;
memset(hash,0,sizeof(hash));
for(i=-50;i<=50;i++)    //枚举前3个；
{
for(j=-50;j<=50;j++)
{
for(k=-50;k<=50;k++)
{
if(i!=0&&j!=0&&k!=0)
{
s=a[1]*i*i*i+a[2]*j*j*j+a[3]*k*k*k;
s=s+MAX;
s=s%MAX;
hash[s]++;        //储存有多少个相同的数；
}
}
}
}
for(q=-50;q<=50;q++)    //枚举后2个；
{
for(w=-50;w<=50;w++)
{
if(q!=0&&w!=0)
{
s=-a[4]*q*q*q-a[5]*w*w*w;
s=s+MAX;                     //hash，相反数模大质数的方法；
sum=sum+hash[s%MAX];
}
}
}
printf("%d\n",sum);
}
int main()
{
int i;
for(i=1;i<=5;i++)
{
scanf("%d",&a[i]);
}
search();
return 0;
}

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