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                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?

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