POJ 1840 Eqs HASH

求一个五元三次方程的解的个数，虽然x的范围比较小，但是要枚举五个数字时间复杂度也是100的五

次方，肯定会超时，将方程改变下，前两项移到左边，加个负号。这样算的时间复杂度就变成了100^3 + 100^2，

不会超时了，值得注意的是HASH数组很大，用short才不会超出内存。

 

/*Accepted    49136 KB    735 ms    C++    1156 B    2012-08-23 09:16:51*/

#include<stdio.h>

#include<string.h>

#include<stdlib.h>



const int MAXN = 25000000;

short hash[MAXN + 1];

int a1, a2, a3, a4, a5;

int l = -50, r = 50;

void DoHash()

{

    int x1, x2, sum;

    memset(hash, 0, sizeof hash);

    for(x1 = l; x1 <= r; x1 ++)

    {

        if(x1 == 0) continue;

        for(x2 = l; x2 <= r; x2 ++)

        {

            if(x2 == 0) continue;

            sum = -(a1 * x1 * x1 * x1 + a2 * x2 * x2 * x2);

            if(sum < 0) sum += MAXN;

            hash[sum] ++;

        }

    }

}



int cal()

{

    int ans = 0, x3, x4, x5, sum;

    for(x3 = l; x3 <= r; x3 ++)

    {

        if(x3 == 0) continue;

        for(x4 = l; x4 <= r; x4 ++)

        {

            if(x4 == 0) continue;

            for(x5 = l; x5 <= r; x5 ++)

            {

                if(x5 == 0) continue;

                sum = a3 * x3 * x3 * x3 + a4 * x4 * x4 * x4 + a5 * x5 * x5 * x5;

                if(sum < 0) sum += MAXN;

                ans += hash[sum];

            }

        }

    }

    return ans;

}



int main()

{

    while(scanf("%d%d%d%d%d", &a1, &a2, &a3, &a4, &a5) != EOF)

    {

        DoHash();

        printf("%d\n", cal());

    }

    return 0;

}