POJ 1840 Eqs（乱搞）题解

思路：这题好像以前有类似的讲过，我们把等式移一下，变成 -（a1*x1^3 + a2*x2^3）== a3*x3^3 + a4*x4^3 + a5*x5^3，那么我们只要先预处理求出左边的答案，然后再找右边是否也能得到就行了，暴力的复杂度从O(n^5)降为O(n^3 + n^2)。因为左式范围-12500000~12500000，所以至少开12500000 * 2的空间，用int会爆，这里用short。如果小于0要加25000000，这样不会有重复的答案，算是简单的hash？

代码：

#include<map>
#include<ctime>
#include<cmath>
#include<cstdio>
#include<iostream>
#include<algorithm>
#define ll long long
#define ull unsigned long long
using namespace std;
const int maxn = 100000 + 10;
const int seed = 131;
const int MOD = 100013;
const int INF = 0x3f3f3f3f;
short num[12500000 * 2 + 10];
int main(){
    int base = 12500000 * 2;
    int a1, a2, a3, a4, a5;
    while(~scanf("%d%d%d%d%d", &a1, &a2, &a3, &a4, &a5)){
        memset(num, 0, sizeof(num));
        for(int i = -50; i <= 50; i++){
            if(!i) continue;
            for(int j = -50; j <= 50; j++){
                if(!j) continue;
                int sum = -(a1 * i * i * i + a2 * j * j * j);
                if(sum < 0) sum += base;
                num[sum]++;
            }
        }
        int ans = 0;
        for(int i = -50; i <= 50; i++){
            if(!i) continue;
            for(int j = -50; j <= 50; j++){
                if(!j) continue;
                for(int k = -50; k <= 50; k++){
                    if(!k) continue;
                    int sum = a3 * i * i * i + a4 * j * j * j + a5 * k * k * k;
                    if(sum < 0) sum += base;
                    ans += num[sum];
                }
            }
        }
        printf("%d
",ans);
    }
    return 0;
}