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;
}