SRM 500(2-1000pt)

DIV2 1000pt

题意：给定两个集合A和B，A = {b1*q1ⁱ | 0 <= i <= n1-1}，B = {b2*q2ⁱ | 0 <= i <= n2-1}，问将AB两个集合合并之后的集合中元素的个数。(注意，每个集合中每个元素只能有一个)。其中0 <= b1,b2,q1,q2 <= 5*10^8，1 <= n1,n2 <= 10^5。

解法：首先，我们称b = 0或者q = 0或者q = 1的集合为特殊集合，因为特殊集合中最多有两个元素。若至少有一个集合为特殊集合，则此问题容易解决。下面考虑两个集合都不为特殊集合的情况。

　　　其实，如果不是A和B中的数太大，我们可以将他们每个都求出来，然后放到一个set<long long>里面，返回set.size()即可，时间复杂度O(n1 + n2)。我们需要找到一种表示这些大数的方法。考虑整数的唯一分解式。

　　　每个整数可以表示成(a1^p1) * (a2^p2) * (a3^p3) *..* (ak^pk)的形式，也就是说，我们只需要统一所有会用到的质数，然后把p1,p2..pk放到一个vector里面，就可以表示每个整数。然后用一个set<vector<long long> >即可统计元素的个数。

tag: math, set

// BEGIN CUT HERE
/*
 * Author:  plum rain
 * score :
 */
/*

 */
// END CUT HERE
#line 11 "GeometricProgressions.cpp"
#include <sstream>
#include <stdexcept>
#include <functional>
#include <iomanip>
#include <numeric>
#include <fstream>
#include <cctype>
#include <iostream>
#include <cstdio>
#include <vector>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <cstdlib>
#include <set>
#include <queue>
#include <bitset>
#include <list>
#include <string>
#include <utility>
#include <map>
#include <ctime>
#include <stack>

using namespace std;

#define CLR(x) memset(x, 0, sizeof(x))
#define PB push_back
#define SZ(v) ((int)(v).size())
#define zero(x) (((x)>0?(x):-(x))<eps)
#define out(x) cout<<#x<<":"<<(x)<<endl
#define tst(a) cout<<#a<<endl
#define CINBEQUICKER std::ios::sync_with_stdio(false)

typedef vector<int> VI;
typedef vector<string> VS;
typedef vector<double> VD;
typedef long long int64;

const double eps = 1e-8;
const double PI = atan(1.0)*4;
const int maxint = 2139062143;
const int N = 10005;
const int M = 10005;

int an[4][N];
int tmp_sz;
int64 bn[4][N];

int inte_dev(int x, int* an, int64* bn)
{
    int all = -1;
    for (int i = 2; i*i <= x;){
        if (!(x%i)){
            an[++all] = i;
            bn[all] = 0;
        }
        while (!(x%i)){
            ++ bn[all];
            x /= i;
        }
        if (i == 2) ++ i;
        else i += 2;
    }
    ++ all;
    if (x != 1){
        an[all] = x; 
        bn[all++] = 1;
    }
    return all;
}

int gao(int64 x, int a)
{
    int ret = 0;
    while (!(x % a)){
        x /= a;
        ++ ret;
    }
    return ret;
}

VI vadd(VI a, VI b)
{
    VI ret; ret.clear();
    for (int i = 0; i < tmp_sz; ++ i)
        ret.PB (a[i]+b[i]);
    return ret;
}

class GeometricProgressions
{
    public:
        int count(int aa, int b, int n, int c, int d, int m){
            int64 a[4];
            a[0] = aa; a[1] = b; a[2] = c; a[3] = d;
            if (!a[2] || !a[3] || a[3] == 1){
                swap (a[0], a[2]); swap (a[1], a[3]); swap (n, m);
            }
            if (!a[0] || !a[1] || a[1] == 1){
                set<int64> tmp;
                tmp.insert(a[0]);
                if (n > 1) tmp.insert(a[1]*a[0]);

                int cnt = 0, sz_tmp = tmp.size();
                int64 now = a[2];
                for (int i = 0; i < m; ++ i){
                    if (tmp.count(now)) ++ cnt;
                    else tmp.insert(now);
                    now *= a[3];
                    if (now > 25e16) 
                        return m + sz_tmp - cnt;
                }
                return tmp.size();
            }

            int64 all[4];
            for (int i = 0; i < 4; ++ i)
                all[i] = inte_dev(a[i], an[i], bn[i]);

            set<int> tmp;
            VI tt; tt.clear();
            for (int i = 0; i < 4; ++ i)
                for (int j = 0; j < all[i]; ++ j)
                    if (!tmp.count(an[i][j])){
                        tmp.insert (an[i][j]);
                        tt.PB (an[i][j]);
                    }

            vector<int> v[4];
            for (int i = 0; i < 4; ++ i)
                v[i].clear();
            tmp_sz = tmp.size();
            for (int i = 0; i < 4; ++ i)
                for (int j = 0; j < tmp_sz; ++ j)
                    v[i].PB (gao(a[i], tt[j]));

            set<VI > ans;
            ans.erase(ans.begin(), ans.end());
            VI now = v[0];
            for (int i = 0; i < n; ++ i){
                if (!ans.count(now)) ans.insert(now);
                now = vadd(now, v[1]);
            }
            now = v[2];
            for (int i = 0; i < m; ++ i){
                if (!ans.count(now)) ans.insert(now);
                now = vadd(now, v[3]);
            }
            return ans.size();
        }
        
// BEGIN CUT HERE
    public:
    //void run_test(int Case) { if ((Case == -1) || (Case == 0)) test_case_0(); if ((Case == -1) || (Case == 1)) test_case_1(); if ((Case == -1) || (Case == 2)) test_case_2(); if ((Case == -1) || (Case == 3)) test_case_3(); }
    void run_test(int Case) { if ((Case == -1) || (Case == 0)) test_case_0();}
    private:
    template <typename T> string print_array(const vector<T> &V) { ostringstream os; os << "{ "; for (typename vector<T>::const_iterator iter = V.begin(); iter != V.end(); ++iter) os << '"' << *iter << "","; os << " }"; return os.str(); }
    void verify_case(int Case, const int &Expected, const int &Received) { cerr << "Test Case #" << Case << "..."; if (Expected == Received) cerr << "PASSED" << endl; else { cerr << "FAILED" << endl; cerr << "	Expected: "" << Expected << '"' << endl; cerr << "	Received: "" << Received << '"' << endl; } }
    void test_case_0() { int Arg0 = 7; int Arg1 = 4; int Arg2 = 5; int Arg3 = 8; int Arg4 = 0; int Arg5 = 1; int Arg6 = 6; verify_case(0, Arg6, count(Arg0, Arg1, Arg2, Arg3, Arg4, Arg5)); }
    void test_case_1() { int Arg0 = 3; int Arg1 = 2; int Arg2 = 5; int Arg3 = 2; int Arg4 = 3; int Arg5 = 5; int Arg6 = 9; verify_case(1, Arg6, count(Arg0, Arg1, Arg2, Arg3, Arg4, Arg5)); }
    void test_case_2() { int Arg0 = 1; int Arg1 = 1; int Arg2 = 1; int Arg3 = 0; int Arg4 = 0; int Arg5 = 1; int Arg6 = 2; verify_case(2, Arg6, count(Arg0, Arg1, Arg2, Arg3, Arg4, Arg5)); }
    void test_case_3() { int Arg0 = 3; int Arg1 = 4; int Arg2 = 100500; int Arg3 = 48; int Arg4 = 1024; int Arg5 = 1000; int Arg6 = 100500; verify_case(3, Arg6, count(Arg0, Arg1, Arg2, Arg3, Arg4, Arg5)); }

// END CUT HERE

};

// BEGIN CUT HERE
int main()
{
//    freopen( "a.out" , "w" , stdout );    
    GeometricProgressions ___test;
    ___test.run_test(-1);
       return 0;
}
// END CUT HERE