POJ 1840 Eqs(hash)

题意  输入a1,a2,a3,a4,a5  求有多少种不同的x1,x2,x3,x4,x5序列使得等式成立   a,x取值在-50到50之间

直接暴力的话肯定会超时的   100的五次方  10e了都    然后能够考虑将等式变一下形   把a1*x1^3+a2*x2^3移到右边   也就是-(a1*x1^3+a2^x2^3)=a3*x3^3+a4*x4^3+a5*x5^3

考虑到a1*x1^3+a2^x2^3的最大值50*50^3+50*50^3=12500000  这个数并不大  能够开这么大的数组把每一个结果出现的次数存下来   又由于结果最小可能是负的12500000   负数不能做数组的下标   加上个12500000*2即可了   这样分别枚举左右两边   把左边出现过的结果都存在一个数组里面    再枚举右边   没出现一次结果  答案就加上前面这个结果出现的次数    枚举完就出现答案了

#include<cstdio>
#include<cstring>
using namespace std;
const int maxs = 50 * 50 * 50 * 50 * 4 + 10;
unsigned short cnt[maxs];
int main()
{
    int a1, a2, a3, a4, a5, sum,ans=0;
    scanf ("%d%d%d%d%d", &a1, &a2, &a3, &a4, &a5);
    for (int x1 = -50; x1 <= 50; ++x1)
    {
        if (x1 == 0) ++x1;
        for (int x2 = -50; x2 <= 50; ++x2)
        {
            if (x2 == 0) ++x2;
            sum = (a1 * x1 * x1 * x1 + a2 * x2 * x2 * x2) * (-1);
            if (sum < 0) ++cnt[sum + maxs];
            else ++cnt[sum];
        }
    }

    for (int x3 = -50; x3 <= 50; ++x3)
    {
        if (x3 == 0) ++x3;
        for (int x4 = -50; x4 <= 50; ++x4)
        {
            if (x4 == 0) ++x4;
            for (int x5 = -50; x5 <= 50; ++x5)
            {
                if (x5 == 0) ++x5;
                sum = (a3 * x3 * x3 * x3 + a4 * x4 * x4 * x4 + a5 * x5 * x5 * x5) ;
                if (sum < 0) sum += maxs;
                ans += cnt[sum];
            }
        }
    }
    printf ("%d\n", ans);
    return 0;
}

Description

Input

Output

Sample Input

37 29 41 43 47

Sample Output

654

还实用hashmap做的  空间优化了不少

#include<iostream>
#include<cstdio>
#include<cstring>
#include<hash_map>
using namespace std;
int first[50*50*50+10];
int ecnt,w[10005],v[10005],nex[10005];

void add(int x)
{
    int t=(x+50*50*50*100)/200,flag=1;
    for(int e=first[t];(~e)&&flag;e=nex[e])
    {
        if(v[e]==x)
        {
            flag=0,w[e]++;
        }
    }
    if(flag)
    {
        w[ecnt]=1;
        v[ecnt]=x;
        nex[ecnt]=first[t];
        first[t]=ecnt++;
    }
}
int getcnt(int x)
{
    if(x>50*50*50*50*2||x<-50*50*50*50*2)return 0;
    int t=(x+50*50*50*100)/200;
    for(int e=first[t];(~e);e=nex[e])
    {
        if(v[e]==x)return w[e];
    }
    return 0;
}
int main()
{
    int a1, a2, a3, a4, a5, sum,ans=0;
    scanf ("%d%d%d%d%d", &a1, &a2, &a3, &a4, &a5);
    memset(first,-1,sizeof first);
    ecnt=0;
    for (int x1 = -50; x1 <= 50; ++x1)
    {
        if (x1 == 0) ++x1;
        for (int x2 = -50; x2 <= 50; ++x2)
        {
            if (x2 == 0) ++x2;
            sum = (a1 * x1 * x1 * x1 + a2 * x2 * x2 * x2) * (-1);
            add(sum);
        }
    }
    for (int x3 = -50; x3 <= 50; ++x3)
    {
        if (x3 == 0) ++x3;
        for (int x4 = -50; x4 <= 50; ++x4)
        {
            if (x4 == 0) ++x4;
            for (int x5 = -50; x5 <= 50; ++x5)
            {
                if (x5 == 0) ++x5;
                sum = (a3 * x3 * x3 * x3 + a4 * x4 * x4 * x4 + a5 * x5 * x5 * x5) ;
                ans +=getcnt(sum);
            }
        }
    }
    printf ("%d\n", ans);
    return 0;
}