poj1840 Eqs

Eqs

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 8835		Accepted: 4362

Description

Consider equations having the following form: 
a1x1 ³+ a2x2 ³+ a3x3 ³+ a4x4 ³+ a5x5 ³=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

Romania OI 2002

这种问一个等式有几种答案的一般都是前几个算一下然后hash，再算后几个，像今年有场多校的一题。用map会超时，要手动hash离散化，查找用二分。

代码

#include <stdio.h>
#include <algorithm>
using namespace std;

int a[10];
int hash[1000005];
int now[1000005];
int sum[1000005];
int upp;

int Count(int x,int y,int z)
{
    return a[0]*x*x*x+a[1]*y*y*y+a[2]*z*z*z;
}

int Find(int t)
{
    int l,r,mid;
    l=0;
    r=upp-1;
    while(l<=r)
    {
        mid=(l+r)/2;
        if (hash[mid]==t) return sum[mid];
        if (hash[mid]<t) l=mid+1;
        else r=mid-1;
    }
    return 0;
}

int main()
{
    int i,j,n,ans,k,tag,up,num,val;
    while(scanf("%d",&a[0])!=EOF)
    {
        up=0;
        for (i=1;i<5;i++)
        {
            scanf("%d",&a[i]);
        }
        for (i=-50;i<=50;i++)
        {
            if (i==0) continue;
            for (j=-50;j<=50;j++)
            {
                if (j==0) continue;
                for (k=-50;k<=50;k++)
                {
                    if (k==0) continue;
                    now[up++]=Count(i,j,k);
                }
            }
        }
        sort(now,now+up);
        upp=0;
        for (i=0;i<up;i++)
        {
            if (i==0)
            {
                num=1;
                val=now[0];
                continue;
            }
            if (val!=now[i])
            {
                hash[upp]=val;
                sum[upp++]=num;
                num=1;
                val=now[i];
            }
            else num++;
        }
        hash[upp]=val;
        sum[upp++]=num;
        ans=0;
        for (i=-50;i<=50;i++)
        {
            if (i==0) continue;
            for (j=-50;j<=50;j++)
            {
                if (j==0) continue;
                tag=i*i*i*a[3]+j*j*j*a[4];
                tag=-tag;
                ans+=Find(tag);
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}