poj 1840 哈希

Eqs

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 14093		Accepted: 6927

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

Source

#include<iostream>    49664K   610MS
#include<cstdio>
#include<cstring>

using namespace std;

short hash[25000001];  //hash[sum]表示值等于sum的的解的个数(多对1映射)

int main()                         //用int会MLE<span id="transmark"></span>
{
    int a1,a2,a3,a4,a5;  //系数
    scanf("%d%d%d%d%d",&a1,&a2,&a3,&a4,&a5);
    {
        memset(hash,0,sizeof(hash));
        for(int x1=-50; x1<=50; x1++)
        {
            if(!x1)
                continue;
            for(int x2=-50; x2<=50; x2++)
            {
                if(!x2)
                    continue;
                int sum=(a1*x1*x1*x1+a2*x2*x2*x2);
                if(sum<0)
                    sum+=25000000;
                hash[sum]++;
            }
        }
        int num=0;
        for(int x3=-50; x3<=50; x3++)
        {
            if(!x3)
                continue;
            for(int x4=-50; x4<=50; x4++)
            {
                if(!x4)
                    continue;
                for(int x5=-50; x5<=50; x5++)
                {
                    if(!x5)
                        continue;
                    int sum=a3*x3*x3*x3+a4*x4*x4*x4+a5*x5*x5*x5;
                    if(sum>12500000||sum<-12500000)  //防止特殊情况发生
						continue;
                    if(sum<0)
                        sum+=25000000;
                    if(hash[sum])
                        num+=hash[sum];
                }
            }
        }
        cout<<num<<endl;
    }
    return 0;
}

#include<iostream>  //  1164K    1704MS
#include<cstdio>
#include<map>

using namespace std;

int main()
{
    map<int,int>Q;
    int a,b,c,d,e;
    scanf("%d%d%d%d%d",&a,&b,&c,&d,&e);
    Q.clear();
    for(int i=-50; i<=50; i++)
    {
        if(!i)
            continue;
        for(int j=-50; j<=50; j++)
        {
            if(!j)
                continue;
			int sum=a*i*i*i+b*j*j*j;
			Q[-sum]++;
        }
    }
    int num=0;
    for(int i=-50;i<=50;i++)
	{
		if(!i)
		continue;
		for(int j=-50;j<=50;j++)
		{
			if(!j)
				continue;
			for(int k=-50;k<=50;k++)
			{
				if(!k)
					continue;
				int sum=c*i*i*i+d*j*j*j+e*k*k*k;
				if(Q.count(sum))
					num+=Q[sum];
			}
		}
	}
	printf("%d\n",num);

}