POJ-1840 Eqs Hash表

不得不说上次看得的这句话是多么对，再差的Hash表都比map好，hash表的查找速度可不是logn能够比的。

首先将5个部分拆成2+3，我们选取2的部分进行hash，然后再进行3重for循环。

动态申请内存还是比不上静态的啊。

代码如下：

动态申请内存：

#include <cstdio>
#include <cstring>
#include <cstdlib>
#define MOD 20003
using namespace std;

int rec[105];

struct Node
{
    int x;
    Node *next;
}e[20003];

void Hash(int key)
{ 
    Node *p = &e[abs(key)%MOD];
    Node *q = new Node;
    q->x = key;
    q->next = p->next;
    p->next = q;
}

int find(int key)
{
    int ans = 0;
    Node *p = &e[abs(key) % MOD];
    while (p != NULL) {
        if (p->x == key) {
            ++ans;
        }
        p = p->next;
    }
    return ans;
}

int main()
{
    int a, b, c, d, e, ans, x; 
    for (int i = -50; i <= 50; ++i) {
        rec[i+50] = i*i*i;
    }
    while (scanf("%d%d%d%d%d", &a, &b, &c, &d, &e) == 5) {
        ans = 0;
        for (int i = -50; i <= 50; ++i) {
            for (int j = -50; j <= 50; ++j) {
                x = -(a * rec[i+50] + b * rec[j+50]);
                if (i != 0 && j != 0) {
                    Hash(x); 
                }
            }
        }
        for (int i = -50; i <= 50; ++i) {
        //    printf("i = %d\n", i);
            for (int j = -50; j <= 50; ++j) {
                for (int k = -50; k <= 50; ++k) {
                    if (i != 0 && j != 0 && k != 0) {
                        x = c * rec[i+50] + d * rec[j+50] + e * rec[k+50];
                        ans += find(x);
                    }
                }
            }
        }
        printf("%d\n", ans);
    }
    return 0;
}

静态：

#include <cstdio>
#include <cstring>
#include <cstdlib>
#define MOD 20003
using namespace std;

int rec[105], idx;

struct Node
{
    int x, cnt, next; 
}e[20003];

int head[20003];

void Hash(int key)
{ 
    int flag = 0, x = key;
    key = abs(key) % MOD;
    for (int i = head[key]; i != -1; i = e[i].next) {
        if (e[i].x == x) {
            flag = 1;
            ++e[i].cnt;
        }
    }
    if (!flag) {
        ++idx;
        e[idx].x = x, e[idx].cnt = 1;
        e[idx].next = head[key];
        head[key] = idx;
    }
}

int find(int key)
{
    int ans = 0, x = key;
    key = abs(key) % MOD; 
    for (int i = head[key]; i != -1; i = e[i].next) {
        if (e[i].x == x) {
            ans += e[i].cnt;
        }
    }
    return ans;
}

int main()
{
    int a, b, c, d, e, ans, x; 
    for (int i = -50; i <= 50; ++i) {
        rec[i+50] = i*i*i;
    }
    while (scanf("%d%d%d%d%d", &a, &b, &c, &d, &e) == 5) {
        memset(head, 0xff, sizeof (head));
        idx = -1;
        ans = 0;
        for (int i = -50; i <= 50; ++i) {
            for (int j = -50; j <= 50; ++j) {
                x = -(a * rec[i+50] + b * rec[j+50]);
                if (i != 0 && j != 0) {
                    Hash(x); 
                }
            }
        }
        for (int i = -50; i <= 50; ++i) {
            for (int j = -50; j <= 50; ++j) {
                for (int k = -50; k <= 50; ++k) {
                    if (i != 0 && j != 0 && k != 0) {
                        x = c * rec[i+50] + d * rec[j+50] + e * rec[k+50];
                        ans += find(x);
                    }
                }
            }
        }
        printf("%d\n", ans);
    }
    return 0;
}