poj1840 哈希

Eqs
Time Limit: 5000MS   Memory Limit: 65536K
Total Submissions: 16229   Accepted: 7965

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
#include<iostream>
#include<cstring>
#include<string>
using namespace std;
short hashs[25000001];
int main()
{
    int a,b,c,d,e;
    while(cin>>a>>b>>c>>d>>e){
    int sb=0;
    memset(hashs,0,sizeof(hashs));
    for(int i=-50;i<=50;i++)
    {
        if(i==0)
            continue;
        for(int j=-50;j<=50;j++)
        {
            if(j==0)
                continue;
            int sum=-(a*i*i*i+b*j*j*j);
            if(sum<0)
                sum+=25000000;
            hashs[sum]++;
        }
    }
        for(int i=-50;i<=50;i++)
        {
            if(i==0)
                continue;
            for(int j=-50;j<=50;j++)
            {
                if(j==0)
                    continue;
                for(int k=-50;k<=50;k++)
                {
                    if(k==0)
                        continue;
                    int sum=c*i*i*i+d*j*j*j+e*k*k*k;
                    if(sum<0)
                        sum+=25000000;
                    if(hashs[sum])
                        sb+=hashs[sum];

                }
            }
        }
        cout<<sb<<endl;
    }


}

number of the solutions for the given equation.

Sample Input

37 29 41 43 47

Sample Output

654

Source



開始的思路中只有時間複雜度爲0(n^5)簡單做法,毫無疑問肯定要超時。沒有思路只好上別人家的博客做做客,原來是用到哈希!而且思路也不是很get到,爲啥自己想就那麼費勁,點播一下就開竅了呢?唉,什麼時候我能夠讓大把的人來我這做做客呢??還是要多練多思考!

推薦這篇博客http://blog.csdn.net/lyy289065406/article/details/6647387裏面講的非常詳細。值得學習代碼優化的思路



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