• POJ - 1840 - Eqs = 思维

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

    题意:求 \(a_1x_1^3+a_2x_2^3+a_3x_3^3+a_4x_4^3+a_5x_5^3=0\) 的整数解,其中所有变量的取值都是 \([-50,50]\) ,且 \(x_i \neq 0\)

    暴力枚举,但是要怎么分两半呢?事实证明是前半部分分2个,后半部分分3个会更好,为什么呢?

    大概是多了一个 \(\log_{2}{100}\)吧,也是差不多7倍常数了。

    前半部分分两个是:
    \(O(n^2\log(n^2)+n^3\log(n^2))\)

    前半部分分三个就白白多了7倍常数,实属逗比。

    可惜POJ用不了unordered_map,待会手写一发hash看看?

    #include<algorithm>
    #include<cmath>
    #include<cstdio>
    #include<cstring>
    #include<iostream>
    #include<map>
    #include<set>
    #include<stack>
    #include<string>
    #include<queue>
    #include<vector>
    using namespace std;
    typedef long long ll;
    
    map<int, int> M;
    
    int main() {
    #ifdef Yinku
        freopen("Yinku.in", "r", stdin);
    #endif // Yinku
        int a1, a2, a3, a4, a5;
        scanf("%d%d%d%d%d", &a1, &a2, &a3, &a4, &a5);
        for(int x1 = -50; x1 <= 50; ++x1) {
            if(x1 == 0)
                continue;
            int p1 = a1 * x1 * x1 * x1;
            for(int x2 = -50; x2 <= 50; ++x2) {
                if(x2 == 0)
                    continue;
                int p2 = a2 * x2 * x2 * x2;
                M[p1 + p2]++;
            }
        }
        ll ans = 0;
        for(int x3 = -50; x3 <= 50; ++x3) {
            if(x3 == 0)
                continue;
            int p3 = a3 * x3 * x3 * x3;
            for(int x4 = -50; x4 <= 50; ++x4) {
                if(x4 == 0)
                    continue;
                int p4 = a4 * x4 * x4 * x4;
                for(int x5 = -50; x5 <= 50; ++x5) {
                    if(x5 == 0)
                        continue;
                    int p5 = a5 * x5 * x5 * x5;
                    map<int, int>::iterator it = M.find(-p3 - p4 - p5);
                    if(it != M.end())
                        ans += it->second;
                }
            }
        }
        printf("%lld\n", ans);
    }

    一个假的哈希,大概就是把它按余数分裂成几棵平衡树来减小树的规模,大概取值合理的话可以快3倍左右(原本平衡树应该是 \(\log_2{10^6}=20\) 的,套个余数哈希(余数为 \(5\times10^4\) )就快了三倍,大概符合 \(\log_2{10^2}=7\) ),注意初始化map是需要时间的,所以并不是余数取越大越好,而且的确会创建map的实例,占用内存空间。

    #include<algorithm>
    #include<cmath>
    #include<cstdio>
    #include<cstring>
    #include<iostream>
    #include<map>
    #include<set>
    #include<stack>
    #include<string>
    #include<queue>
    #include<vector>
    using namespace std;
    typedef long long ll;
    
    const int MAXN = 49999;
    struct HashTable {
        map<int, int> M[MAXN];
        void insert(int x) {
            int p = x % MAXN;
            if(p < 0)
                p += MAXN;
            M[p][x]++;
        }
        int count(int x) {
            int p = x % MAXN;
            if(p < 0)
                p += MAXN;
            map<int, int>::iterator it = M[p].find(x);
            if(it != M[p].end())
                return it->second;
            return 0;
        }
    } ht;
    
    //寻找n以内的一个最大的质数
    /*const int MAXP=2e6;
    bool np[MAXP+1];
    void find_p(int n){
        np[1]=1;
        for(int i=1;i<=n;++i){
            if(np[i])
                continue;
            for(int j=i+i;j<=n;j+=i)
                np[j]=1;
        }
        for(int i=n;;--i){
            if(!np[i]){
                printf("MAXP=%d\n",i);
                break;
            }
        }
    }*/
    
    int main() {
    #ifdef Yinku
        freopen("Yinku.in", "r", stdin);
    #endif // Yinku
        //find_p(5e4);
        int a1, a2, a3, a4, a5;
        scanf("%d%d%d%d%d", &a1, &a2, &a3, &a4, &a5);
        for(int x1 = -50; x1 <= 50; ++x1) {
            if(x1 == 0)
                continue;
            int p1 = a1 * x1 * x1 * x1;
            for(int x2 = -50; x2 <= 50; ++x2) {
                if(x2 == 0)
                    continue;
                int p2 = a2 * x2 * x2 * x2;
                ht.insert(p1 + p2);
            }
        }
        ll ans = 0;
        for(int x3 = -50; x3 <= 50; ++x3) {
            if(x3 == 0)
                continue;
            int p3 = a3 * x3 * x3 * x3;
            for(int x4 = -50; x4 <= 50; ++x4) {
                if(x4 == 0)
                    continue;
                int p4 = a4 * x4 * x4 * x4;
                for(int x5 = -50; x5 <= 50; ++x5) {
                    if(x5 == 0)
                        continue;
                    int p5 = a5 * x5 * x5 * x5;
                    ans += ht.count(-p3 - p4 - p5);
                }
            }
        }
        printf("%lld\n", ans);
    }

    但是假如哈希套哈希再套平衡树说不定会快到飞起?

    相关文章
    相关标签/搜索
    最准网站特马资料四肖期期准一2020特马开奖结果查询三肖中特期期准免费118图库