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

  1 // BEGIN CUT HERE
  2 /*
  3  * Author:  plum rain
  4  * score :
  5  */
  6 /*
  7 
  8  */
  9 // END CUT HERE
 10 #line 11 "GeometricProgressions.cpp"
 11 #include <sstream>
 12 #include <stdexcept>
 13 #include <functional>
 14 #include <iomanip>
 15 #include <numeric>
 16 #include <fstream>
 17 #include <cctype>
 18 #include <iostream>
 19 #include <cstdio>
 20 #include <vector>
 21 #include <cstring>
 22 #include <cmath>
 23 #include <algorithm>
 24 #include <cstdlib>
 25 #include <set>
 26 #include <queue>
 27 #include <bitset>
 28 #include <list>
 29 #include <string>
 30 #include <utility>
 31 #include <map>
 32 #include <ctime>
 33 #include <stack>
 34 
 35 using namespace std;
 36 
 37 #define CLR(x) memset(x, 0, sizeof(x))
 38 #define PB push_back
 39 #define SZ(v) ((int)(v).size())
 40 #define zero(x) (((x)>0?(x):-(x))<eps)
 41 #define out(x) cout<<#x<<":"<<(x)<<endl
 42 #define tst(a) cout<<#a<<endl
 43 #define CINBEQUICKER std::ios::sync_with_stdio(false)
 44 
 45 typedef vector<int> VI;
 46 typedef vector<string> VS;
 47 typedef vector<double> VD;
 48 typedef long long int64;
 49 
 50 const double eps = 1e-8;
 51 const double PI = atan(1.0)*4;
 52 const int maxint = 2139062143;
 53 const int N = 10005;
 54 const int M = 10005;
 55 
 56 int an[4][N];
 57 int tmp_sz;
 58 int64 bn[4][N];
 59 
 60 int inte_dev(int x, int* an, int64* bn)
 61 {
 62     int all = -1;
 63     for (int i = 2; i*i <= x;){
 64         if (!(x%i)){
 65             an[++all] = i;
 66             bn[all] = 0;
 67         }
 68         while (!(x%i)){
 69             ++ bn[all];
 70             x /= i;
 71         }
 72         if (i == 2) ++ i;
 73         else i += 2;
 74     }
 75     ++ all;
 76     if (x != 1){
 77         an[all] = x; 
 78         bn[all++] = 1;
 79     }
 80     return all;
 81 }
 82 
 83 int gao(int64 x, int a)
 84 {
 85     int ret = 0;
 86     while (!(x % a)){
 87         x /= a;
 88         ++ ret;
 89     }
 90     return ret;
 91 }
 92 
 93 VI vadd(VI a, VI b)
 94 {
 95     VI ret; ret.clear();
 96     for (int i = 0; i < tmp_sz; ++ i)
 97         ret.PB (a[i]+b[i]);
 98     return ret;
 99 }
100 
101 class GeometricProgressions
102 {
103     public:
104         int count(int aa, int b, int n, int c, int d, int m){
105             int64 a[4];
106             a[0] = aa; a[1] = b; a[2] = c; a[3] = d;
107             if (!a[2] || !a[3] || a[3] == 1){
108                 swap (a[0], a[2]); swap (a[1], a[3]); swap (n, m);
109             }
110             if (!a[0] || !a[1] || a[1] == 1){
111                 set<int64> tmp;
112                 tmp.insert(a[0]);
113                 if (n > 1) tmp.insert(a[1]*a[0]);
114 
115                 int cnt = 0, sz_tmp = tmp.size();
116                 int64 now = a[2];
117                 for (int i = 0; i < m; ++ i){
118                     if (tmp.count(now)) ++ cnt;
119                     else tmp.insert(now);
120                     now *= a[3];
121                     if (now > 25e16) 
122                         return m + sz_tmp - cnt;
123                 }
124                 return tmp.size();
125             }
126 
127             int64 all[4];
128             for (int i = 0; i < 4; ++ i)
129                 all[i] = inte_dev(a[i], an[i], bn[i]);
130 
131             set<int> tmp;
132             VI tt; tt.clear();
133             for (int i = 0; i < 4; ++ i)
134                 for (int j = 0; j < all[i]; ++ j)
135                     if (!tmp.count(an[i][j])){
136                         tmp.insert (an[i][j]);
137                         tt.PB (an[i][j]);
138                     }
139 
140             vector<int> v[4];
141             for (int i = 0; i < 4; ++ i)
142                 v[i].clear();
143             tmp_sz = tmp.size();
144             for (int i = 0; i < 4; ++ i)
145                 for (int j = 0; j < tmp_sz; ++ j)
146                     v[i].PB (gao(a[i], tt[j]));
147 
148             set<VI > ans;
149             ans.erase(ans.begin(), ans.end());
150             VI now = v[0];
151             for (int i = 0; i < n; ++ i){
152                 if (!ans.count(now)) ans.insert(now);
153                 now = vadd(now, v[1]);
154             }
155             now = v[2];
156             for (int i = 0; i < m; ++ i){
157                 if (!ans.count(now)) ans.insert(now);
158                 now = vadd(now, v[3]);
159             }
160             return ans.size();
161         }
162         
163 // BEGIN CUT HERE
164     public:
165     //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(); }
166     void run_test(int Case) { if ((Case == -1) || (Case == 0)) test_case_0();}
167     private:
168     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(); }
169     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; } }
170     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)); }
171     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)); }
172     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)); }
173     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)); }
174 
175 // END CUT HERE
176 
177 };
178 
179 // BEGIN CUT HERE
180 int main()
181 {
182 //    freopen( "a.out" , "w" , stdout );    
183     GeometricProgressions ___test;
184     ___test.run_test(-1);
185        return 0;
186 }
187 // END CUT HERE
View Code

 

posted @ 2013-12-07 10:38  Plumrain  阅读(249)  评论(0)  编辑  收藏  举报