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-;
 const double PI = atan(1.0)*;
 const int maxint = ;
 const int N = ;
 const int M = ;
 
 int an[][N];
 int tmp_sz;
 int64 bn[][N];
 
 int inte_dev(int x, int* an, int64* bn)
 {
     int all = -;
     for (int i = ; i*i <= x;){
         if (!(x%i)){
             an[++all] = i;
             bn[all] = ;
         }
         while (!(x%i)){
             ++ bn[all];
             x /= i;
         }
         if (i == ) ++ i;
         else i += ;
     }
     ++ all;
     if (x != ){
         an[all] = x;
         bn[all++] = ;
     }
     return all;
 }
 
 int gao(int64 x, int a)
 {
     int ret = ;
     while (!(x % a)){
         x /= a;
         ++ ret;
     }
     return ret;
 }
 
 VI vadd(VI a, VI b)
 {
     VI ret; ret.clear();
     for (int i = ; 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[];
             a[] = aa; a[] = b; a[] = c; a[] = d;
             if (!a[] || !a[] || a[] == ){
                 swap (a[], a[]); swap (a[], a[]); swap (n, m);
             }
             if (!a[] || !a[] || a[] == ){
                 set<int64> tmp;
                 tmp.insert(a[]);
                 if (n > ) tmp.insert(a[]*a[]);
 
                 int cnt = , sz_tmp = tmp.size();
                 int64 now = a[];
                 for (int i = ; i < m; ++ i){
                     if (tmp.count(now)) ++ cnt;
                     else tmp.insert(now);
                     now *= a[];
                     if (now > 25e16)
                         return m + sz_tmp - cnt;
                 }
                 return tmp.size();
             }
 
             int64 all[];
             for (int i = ; i < ; ++ i)
                 all[i] = inte_dev(a[i], an[i], bn[i]);
 
             set<int> tmp;
             VI tt; tt.clear();
             for (int i = ; i < ; ++ i)
                 for (int j = ; j < all[i]; ++ j)
                     if (!tmp.count(an[i][j])){
                         tmp.insert (an[i][j]);
                         tt.PB (an[i][j]);
                     }
 
             vector<int> v[];
             for (int i = ; i < ; ++ i)
                 v[i].clear();
             tmp_sz = tmp.size();
             for (int i = ; i < ; ++ i)
                 for (int j = ; j < tmp_sz; ++ j)
                     v[i].PB (gao(a[i], tt[j]));
 
             set<VI > ans;
             ans.erase(ans.begin(), ans.end());
             VI now = v[];
             for (int i = ; i < n; ++ i){
                 if (!ans.count(now)) ans.insert(now);
                 now = vadd(now, v[]);
             }
             now = v[];
             for (int i = ; i < m; ++ i){
                 if (!ans.count(now)) ans.insert(now);
                 now = vadd(now, v[]);
             }
             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 == -) || (Case == )) 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 << "\tExpected: \"" << Expected << '\"' << endl; cerr << "\tReceived: \"" << Received << '\"' << endl; } }
     void test_case_0() { int Arg0 = ; int Arg1 = ; int Arg2 = ; int Arg3 = ; int Arg4 = ; int Arg5 = ; int Arg6 = ; verify_case(, Arg6, count(Arg0, Arg1, Arg2, Arg3, Arg4, Arg5)); }
     void test_case_1() { int Arg0 = ; int Arg1 = ; int Arg2 = ; int Arg3 = ; int Arg4 = ; int Arg5 = ; int Arg6 = ; verify_case(, Arg6, count(Arg0, Arg1, Arg2, Arg3, Arg4, Arg5)); }
     void test_case_2() { int Arg0 = ; int Arg1 = ; int Arg2 = ; int Arg3 = ; int Arg4 = ; int Arg5 = ; int Arg6 = ; verify_case(, Arg6, count(Arg0, Arg1, Arg2, Arg3, Arg4, Arg5)); }
     void test_case_3() { int Arg0 = ; int Arg1 = ; int Arg2 = ; int Arg3 = ; int Arg4 = ; int Arg5 = ; int Arg6 = ; verify_case(, 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(-);
        return ;
 }
 // END CUT HERE