SRM 409(1-250pt, 1-500pt)

DIV1 250pt

题意：称string s是vector<string> words的ordered superstring，如果它满足：存在一个数列{x0, x1, x2...xm}（m = words.size()），使得words[i]与s中从xi开始的，长度为words[i].size()的字符串相同，且x0 <= x1 <= x2 <= ... <= xm。

　　　给定words，求最短的ordered superstring。words.size() <= 50，words[i].size() <= 50。

解法：模拟题。由于使用指针比较易错加上我代码太不稳于是wa了。。。。

tag:simulation

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

  */
 // END CUT HERE
 #line 11 "OrderedSuperString.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 clr0(x) memset(x, 0, sizeof(x))
 #define clr1(x) memset(x, -1, sizeof(x))
 #define pb push_back
 #define sz(v) ((int)(v).size())
 #define all(t) t.begin(),t.end()
 #define zero(x) (((x)>0?(x):-(x))<eps)
 #define out(x) cout<<#x<<":"<<(x)<<endl
 #define tst(a) cout<<a<<" "
 #define tst1(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 pair<int, int> pii;
 typedef long long int64;

 const double eps = 1e-;
 const double PI = atan(1.0)*;
 const int inf =  / ;

 bool ok(string a, string b)
 {
     for (int i = ; i < min(sz(a), sz(b)); ++ i)
         if (a[i] != b[i]) return ;
     return ;
 }

 class OrderedSuperString
 {
     public:
         int getLength(vector <string> w){
             string s; s.clear();
             int idx = ;
             for (int i = ; i < sz(w); ++ i){
                 int match = sz(s);
                 for (int j = idx; j < sz(s); ++ j)
                     if (ok(string(s.begin()+j, s.end()), w[i])){
                         match = j; break;
                     }
                 idx = match;
                 if (sz(s) - match < sz(w[i]))
                     s += string(w[i].begin()+sz(s)-match, w[i].end());
             }
             return sz(s);
         }

 // BEGIN CUT HERE
     public:
     void run_test(int Case) { if ((Case == -) || (Case == )) test_case_0(); if ((Case == -) || (Case == )) test_case_1(); if ((Case == -) || (Case == )) test_case_2(); if ((Case == -) || (Case == )) test_case_3(); }
     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() { string Arr0[] = {"aaaaaaaaaaabaaaaaaaa", "bac", "aaaabacaaa", "ab", "ba", "a", "ca"}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arg1 = ; verify_case(, Arg1, getLength(Arg0)); }
     void test_case_1() { string Arr0[] = {"a","a","b","a"}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arg1 = ; verify_case(, Arg1, getLength(Arg0)); }
     void test_case_2() { string Arr0[] = {"abcdef", "ab","bc", "de","ef"}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arg1 = ; verify_case(, Arg1, getLength(Arg0)); }
     void test_case_3() { string Arr0[] = {"ab","bc", "de","ef","abcdef"}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arg1 = ; verify_case(, Arg1, getLength(Arg0)); }

 // END CUT HERE

 };

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

DIV1 600pt

题意：给3个正整数n1, n2, up，求能被C(n1+n2, n1)整除的，且小等于up的最大正整数。n1,n2 <= 10^9，up <= 10^5。

解法：首先，枚举小等于up的所有数是肯定的。那么现在考虑，对整数k，如何判定它能不能被C(n1+n2, n1)整数。

　　　当数据太大不能被直接表示，又不能用余数间接表示的时候，就考虑它的所有质因子。只用考虑sqrt(k)以内的所有质因子，或者k为质数。

　　　然后，考虑到组合数的特殊性，C(n1+n2, n1) = (n1+n2)! / (n1! * n2!)。所以下面考虑对某个质因子t，如何求n!含有多少个质因子t。

　　　下图为13!含有因子2的个数，一个X代表一个。也即是说，13的阶乘含有因子2的数量为13/2 + 13/(2^2) + 13/(2^3)。

　　　

tag:math, number theory, good

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

  */
 // END CUT HERE
 #line 11 "MagicalSpheres.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 clr0(x) memset(x, 0, sizeof(x))
 #define clr1(x) memset(x, -1, sizeof(x))
 #define pb push_back
 #define sz(v) ((int)(v).size())
 #define all(t) t.begin(),t.end()
 #define zero(x) (((x)>0?(x):-(x))<eps)
 #define out(x) cout<<#x<<":"<<(x)<<endl
 #define tst(a) cout<<a<<" "
 #define tst1(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 pair<int, int> pii;
 typedef long long int64;

 const double eps = 1e-;
 const double PI = atan(1.0)*;
 const int inf =  / ;
 const int N = ;

 class MagicalSpheres
 {
     public:
         int64 gao (int x, int m)
         {
             int64 num = ;
             int64 tmp = m;
             while (tmp <= x){
                 num += x / tmp;
                 tmp *= m;
             }
             return num;
         }

         int divideWork(int n1, int n2, int up){
             for (int i = up; i; -- i){
                 bool ok = ;
                 int k = i;
                 for (int j = ; j*j <= k; ++ j) if (k % j == ){
                     int t = ;
                     while (!(k % j))
                         k /= j, ++ t;
                     if (gao(n1+n2, j) - gao(n1, j) - gao(n2, j) < t) ok = ;
                 }
                 if (k != )
                     if (gao(n1+n2, k) - gao(n1, k) - gao(n2, k) < ) ok = ;
                 if (!ok) return i;
             }
             return ;
         }

         // BEGIN CUT HERE
     public:
         void run_test(int Case) { if ((Case == -) || (Case == )) test_case_0(); if ((Case == -) || (Case == )) test_case_1(); if ((Case == -) || (Case == )) test_case_2(); if ((Case == -) || (Case == )) test_case_3(); }
     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 = ; verify_case(, Arg3, divideWork(Arg0, Arg1, Arg2)); }
         void test_case_1() { int Arg0 = ; int Arg1 = ; int Arg2 = ; int Arg3 = ; verify_case(, Arg3, divideWork(Arg0, Arg1, Arg2)); }
         void test_case_2() { int Arg0 = ; int Arg1 = ; int Arg2 = ; int Arg3 = ; verify_case(, Arg3, divideWork(Arg0, Arg1, Arg2)); }
         void test_case_3() { int Arg0 = ; int Arg1 = ; int Arg2 = ; int Arg3 = ; verify_case(, Arg3, divideWork(Arg0, Arg1, Arg2)); }

         // END CUT HERE

 };

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