POJ 1840 Eqs (哈希)

Eqs

 

Time Limit: 5000MS Memory Limit: 65536K
Total Submissions: 16200 Accepted: 7953

 

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

 

题意

给出 a1−a5 求满足上式方程的解的组数。

 

思路

因为给出了 xi 的范围,所以如果直接暴力的话用五重循环一定会超时,所以我们可以把这个方程分成左右两边,也就是移项。对左边枚举 xi 取各个值时所求得数值的次数,然后枚举右边,若计算的结果刚好之前已经出现过,则找到了 hash[temp] 组解,求和即可。

 

AC 代码

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<iostream>
using namespace std;
#include<queue>
#include<map>

#define MAXN 25000010

typedef __int64 LL;

int a1,a2,a3,a4,a5;
short hash[MAXN];
int main()
{
    while(~scanf("%d%d%d%d%d",&a1,&a2,&a3,&a4,&a5))
    {
        int ans=0,temp;
        for(int i=-50; i<=50; i++)
            for(int j=-50; j<=50; j++)
            {
                if(i==0||j==0)continue;
                temp=-(a1*i*i*i+a2*j*j*j);
                if(temp<0)temp+=MAXN;
                hash[temp]++;
            }
        for(int i=-50; i<=50; i++)
            for(int j=-50; j<=50; j++)
                for(int k=-50; k<=50; k++)
                {
                    if(i==0||j==0||k==0)continue;
                    temp=a3*i*i*i+a4*j*j*j+a5*k*k*k;
                    if(temp<0)temp+=MAXN;
                    if(hash[temp])
                        ans+=hash[temp];
                }
        printf("%d\n",ans);
    }
    return 0;
}