poj1840 hash

Eqs

题意：给出a1, a2, a3, a4, a5

求满足表达式 a1x1³+ a2x2³+ a3x3³+ a4x4³+ a5x5³=0 的解(x1, x2, x3, x4, x5)的个数，其中xi范围是[-50, 50], xi != 0 (i = 1, 2, 3, 4, 5) 
分析：这题挺简单的，将解分成两组(x1, x2, x3) 和 (x4, x5)，先三重循环求出a1x1³+ a2x2³+ a3x3³他们的解，用开散列法对应到数组中，再用两重循环求出a4x4³+ a5x5³求得结果在hash表中查找然后判断就行了。

#include<stdio.h>
#include<string.h>
#include <iostream>
using namespace std;
#define NN 1000020
#define MAXHASH 100003 //这是一个素数
int x1, x2, x3, x4, x5;
int a1, a2, a3, a4, a5;
int idx;

typedef struct po{
    int sum;
    struct po *nxt;

}PO;

PO f[NN];
PO *head[MAXHASH];

int cube(int x){
    return x * x * x;
}

int hash(int x){
    int h = x % MAXHASH;
    if(h < 0) h += MAXHASH;
    return h;
}
int OK(int tmp){
    int ans = 0;
    int h = hash(tmp);
    for (PO *p = head[h]; p; p = p->nxt){
        if(p->sum == tmp){
            ans++;
        }
    }
    return ans;
}

int main() {
    int tmp, h;
    scanf("%d%d%d%d%d", &a1, &a2, &a3, &a4, &a5);
    idx = 0;
    memset(head, 0, sizeof(head));
    for (x1 = -50; x1 <= 50; x1++){
        if(x1 == 0) continue;
        for (x2 = -50; x2 <= 50; x2++){
            if(x2 == 0) continue;
            for (x3 = -50; x3 <= 50; x3++){
                if(x3 == 0) continue;
                tmp = a1 * cube(x1) + a2 * cube(x2) + a3 * cube(x3);
                h = hash(tmp);
                f[idx].sum = tmp;
                f[idx].nxt = head[h];
                head[h] = f + idx++;
            }
        }
    }
    int res = 0;
    for (x4 = -50; x4 <= 50; x4++){
        if(x4 == 0) continue;
        for (x5 = -50; x5 <= 50; x5++){
            if(x5 == 0) continue;
            tmp = a4 * cube(x4) + a5 * cube(x5);
            tmp = -tmp;
            res += OK(tmp);
        }
    }
    printf("%d\n", res);
    return 0;
}