POJ 1840 Eqs（暴力）

POJ 1840 Eqs（暴力）

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

Romania OI 2002

裸的暴力題，比賽的時候開了5個循環，看時間限制在5秒，以為呀，結果本地都跑不過來。

第一次用short的說。

#include
#include
#include
#include
#include
typedef long long LL;
using namespace std;                        
short hash[25000001];
int a1,a2,a3,a4,a5;

int main()
{
    while(~scanf("%d%d%d%d%d",&a1,&a2,&a3,&a4,&a5))
    {
        int ans=0;
        memset(hash,0,sizeof(hash));
        for(int x1=-50;x1<=50;x1++)
        {
            if(x1==0)
               continue;
            for(int x2=-50;x2<=50;x2++)
            {
                if(x2==0)
                    continue;
                int s=(-1)*(a1*x1*x1*x1+a2*x2*x2*x2);
                if(s<0)
                    s+=25000000;
                hash[s]++;
            }
        }
        for(int x1=-50;x1<=50;x1++)
        {
            if(x1==0)
               continue;
            for(int x2=-50;x2<=50;x2++)
            {
                if(x2==0)
                    continue;
                for(int x3=-50;x3<=50;x3++)
                {
                    if(x3==0)
                        continue;
                    int s=a3*x1*x1*x1+a4*x2*x2*x2+a5*x3*x3*x3;
                    if(s<0)
                        s+=25000000;
                    if(hash[s])
                        ans+=hash[s];
                }
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}