poj 1840 Eqs

http://poj.org/problem?id=1840

　　简单hash。。。。很想知道那些200ms以内的是怎么搞的。。。

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int mod = 4000037;
const int HASH = 0x20decade;
const int inf = 0x7fffffff;

int hash[mod];
int used[mod] = {0};
int tri[101];

void pre() {
    for (int i = -50; i <= 50; i++) {
        tri[i + 50] = i * i * i;
    }
}

void insert(int _x) {
    int p = ((_x ^ HASH) & inf) % mod;

    while (hash[p] != _x && used[p]) p = (p + 1) % mod;
    hash[p] = _x;
    used[p]++;
}

int find(int _x) {
    int p = ((_x ^ HASH) & inf) % mod;

    while (hash[p] != _x && used[p]) p = (p + 1) % mod;

    return used[p];
}

int main() {
    int a[5];
    bool nz = false;

    pre();
    for (int i = 0; i < 5; i++) {
        scanf("%d", &a[i]);
        if (a[i]) nz = true;
    }
    if (!nz) {
        puts("10000000000");

        return 0;
    }
    for (int i = 0; i <= 100; i++) {
        if (i == 50) continue;
        int t1 = tri[i] * a[0];

        for (int j = 0; j <= 100; j++) {
            if (j == 50) continue;
            int t2 = tri[j] * a[1];

            for (int k = 0; k <= 100; k++) {
                if (k == 50) continue;
                insert(t1 + t2 + tri[k] * a[2]);
            }
        }
    }
    __int64 cnt = 0;

    for (int i = 0; i <= 100; i++) {
         if (i == 50) continue;
         int t = tri[i] * a[3];

         for (int j = 0; j <= 100; j++) {
             if (j == 50) continue;
             cnt += find(- (t + tri[j] * a[4]));
         }
    }

    printf("%I64d\n", cnt++);

    return 0;
}

——written by Lyon