DIV1 250pt

题意：给定一个正整数n(n <= 10^18)，如果n = p^q，其中p为质数，q > 1，则返回vector<int> ans = {p, q}，否则返回ans =  {}。

解法：首先，看到题的第一想法是，枚举p。而由于n的范围太大，O(10^9)的复杂度不能接受。又想到，如果题目的限制为q > 2，那么枚举的复杂度就降到了O(10^6)，可以接受了。故得到算法，特判n可不可以表示成p ^ 2的形式(注意判定p是不是质数)，然后再从1到10^6枚举看n能不能表示成p^q，其中q > 2。这样，问题能够得到解决，但是编码复杂度比较高，最后我写出来只拿到了160+分。

　　　然后看了题解，枚举q...因为10^18 < 2^60，所以q的范围小于60。然后枚举就行了。注意三点，一、枚举的方法可以直接用pow来对n开方，也可以二分，前者会涉及浮点数但是好写。二、枚举的过程中算多少次方不需要快速幂，直接算容易写而且时间复杂度够。三、计算过程中要注意会不会爆long long，如果快爆long long直接返回-1或者0。

tag:math

解法一：

  1.  // BEGIN CUT HERE
  2.  /*
  3.   * Author:  plum rain
  4.   * score :
  5.   */
  6.  /*
  7.  
  8.   */
  9.  // END CUT HERE
  10.  #line 11 "StrongPrimePower.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 CLR1(x) memset(x, -1, sizeof(x))
  39.  #define PB push_back
  40.  #define SZ(v) ((int)(v).size())
  41.  #define zero(x) (((x)>0?(x):-(x))<eps)
  42.  #define out(x) cout<<#x<<":"<<(x)<<endl
  43.  #define tst(a) cout<<#a<<endl
  44.  #define CINBEQUICKER std::ios::sync_with_stdio(false)
  45.  
  46.  typedef vector<int> VI;
  47.  typedef vector<string> VS;
  48.  typedef vector<double> VD;
  49.  typedef pair<int, int> pii;
  50.  typedef long long int64;
  51.  
  52.  const double eps = 1e-;
  53.  const double PI = atan(1.0)*;
  54.  const int maxint = ;
  55.  const int maxx = ;
  56.  
  57.  int all;
  58.  bool vis[maxx];
  59.  int prm[maxx];
  60.  
  61.  void sieve(int n)
  62.  {
  63.      int m = (int)sqrt(n+0.5);
  64.      CLR (vis); vis[] = vis[] = ;
  65.      for (int64 i = ; i <= m; ++ i) if (!vis[i])
  66.          for (int64 j = i*i; j <= n; j += i) vis[j] = ;
  67.  
  68.  }
  69.  
  70.  int primes(int n)
  71.  {
  72.      sieve(n);
  73.      int ret = ;
  74.      for (int i = ; i <= n; ++ i)
  75.          if (!vis[i]) prm[ret++] = i;
  76.      return ret;
  77.  
  78.  }
  79.  
  80.  bool ok(int64 n)
  81.  {
  82.      if (n <= ) return (!vis[n]);
  83.  
  84.      for (int i = ; i < all; ++ i)
  85.          if ((n % prm[i]) == ){
  86.              return ;
  87.          }
  88.      return ;
  89.  }
  90.  
  91.  int gao1(int64 n)
  92.  {
  93.      int64 m = (int64)sqrt(n + 0.5);
  94.      if (m*m != n) return -;
  95.  
  96.      if (ok(m)) return m;
  97.      return -;
  98.  }
  99.  
  100.  pii gao2(int64 n)
  101.  {
  102.      int ret;
  103.      pii tmp;
  104.      for (int i = ; i < all; ++ i) if ((n % prm[i]) == ){
  105.          ret = ;
  106.          while ((n % prm[i]) == ) {
  107.              ++ ret;
  108.              n /= prm[i];
  109.          }
  110.          if (ret >  && n == ){
  111.              tmp.first = prm[i];
  112.              tmp.second = ret;
  113.              return tmp;
  114.          }
  115.          else{
  116.              tmp.first = -;
  117.              return tmp;
  118.          }
  119.      }
  120.      tmp.first = -;
  121.      return tmp;
  122.  }
  123.  
  124.  int64 gao(string n)
  125.  {
  126.      int64 ret = ;
  127.      for (int i = ; i < n.size(); ++ i)
  128.          ret = ret *  + n[i] - '';
  129.      return ret;
  130.  }
  131.  
  132.  class StrongPrimePower
  133.  {
  134.      public:
  135.          vector <int> baseAndExponent(string N){
  136.              all = primes();
  137.              int64 n = gao(N);
  138.  
  139.              int64 tmp = gao1(n);
  140.              vector<int> ans;
  141.              ans.clear();
  142.              if (tmp != -){
  143.                  ans.PB ((int)tmp); ans.PB ();
  144.                  return ans;
  145.              }
  146.              pii t = gao2(n);
  147.              if (t.first != -){
  148.                  ans.PB (t.first); ans.PB (t.second);
  149.                  return ans;
  150.              }
  151.              return ans;
  152.          }
  153.  
  154.  // BEGIN CUT HERE
  155.      public:
  156.      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(); if ((Case == -) || (Case == )) test_case_4(); if ((Case == -) || (Case == )) test_case_5(); }
  157.      private:
  158.      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(); }
  159.      void verify_case(int Case, const vector <int> &Expected, const vector <int> &Received) { cerr << "Test Case #" << Case << "..."; if (Expected == Received) cerr << "PASSED" << endl; else { cerr << "FAILED" << endl; cerr << "\tExpected: " << print_array(Expected) << endl; cerr << "\tReceived: " << print_array(Received) << endl; } }
  160.      void test_case_0() { string Arg0 = ""; int Arr1[] = {,  }; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[]))); verify_case(, Arg1, baseAndExponent(Arg0)); }
  161.      void test_case_1() { string Arg0 = ""; int Arr1[] = { }; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[]))); verify_case(, Arg1, baseAndExponent(Arg0)); }
  162.      void test_case_2() { string Arg0 = ""; int Arr1[] = { }; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[]))); verify_case(, Arg1, baseAndExponent(Arg0)); }
  163.      void test_case_3() { string Arg0 = ""; int Arr1[] = { }; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[]))); verify_case(, Arg1, baseAndExponent(Arg0)); }
  164.      void test_case_4() { string Arg0 = ""; int Arr1[] = {,  }; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[]))); verify_case(, Arg1, baseAndExponent(Arg0)); }
  165.      void test_case_5() { string Arg0 = ""; int Arr1[] = {,  }; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[]))); verify_case(, Arg1, baseAndExponent(Arg0)); }
  166.  
  167.  // END CUT HERE
  168.  
  169.  };
  170.  
  171.  // BEGIN CUT HERE
  172.  int main()
  173.  {
  174.  //    freopen( "a.out" , "w" , stdout );
  175.      StrongPrimePower ___test;
  176.      ___test.run_test(-);
  177.         return ;
  178.  }
  179.  // END CUT HERE

解法二：

  1.  // BEGIN CUT HERE
  2.  /*
  3.   * Author:  plum rain
  4.   * score :
  5.   */
  6.  /*
  7.  
  8.   */
  9.  // END CUT HERE
  10.  #line 11 "StrongPrimePower.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 CLR1(x) memset(x, -1, sizeof(x))
  39.  #define PB push_back
  40.  #define SZ(v) ((int)(v).size())
  41.  #define zero(x) (((x)>0?(x):-(x))<eps)
  42.  #define out(x) cout<<#x<<":"<<(x)<<endl
  43.  #define tst(a) cout<<#a<<endl
  44.  #define CINBEQUICKER std::ios::sync_with_stdio(false)
  45.  
  46.  typedef vector<int> VI;
  47.  typedef vector<string> VS;
  48.  typedef vector<double> VD;
  49.  typedef pair<int, int> pii;
  50.  typedef long long int64;
  51.  
  52.  const double eps = 1e-;
  53.  const double PI = atan(1.0)*;
  54.  const int maxint = ;
  55.  
  56.  int64 gao(string n)
  57.  {
  58.      int64 ret = ;
  59.      for (int i = ; i < n.size(); ++ i)
  60.          ret = ret *  + n[i] - '';
  61.      return ret;
  62.  }
  63.  
  64.  int64 mypow(int64 p, int64 n)
  65.  {
  66.      unsigned long long ret = ;
  67.      for (int i = ; i < n; ++ i){
  68.          ret *= p;
  69.          if (ret > 1e18) return -;
  70.      }
  71.      return ret;
  72.  }
  73.  
  74.  bool ok(int y)
  75.  {
  76.      for (int64 i = ; i*i <= y; ++ i)
  77.          if (y % i == ) return ;
  78.      return ;
  79.  }
  80.  
  81.  int gao(int64 n, int num)
  82.  {
  83.      int x = (int)pow(n, 1.0 / (double)num), y = -;
  84.      for (int i = -; i < ; ++ i)
  85.          if (mypow(x+i,num) == n && ok(x+i)) y = x + i;
  86.      return y;
  87.  }
  88.  
  89.  class StrongPrimePower
  90.  {
  91.      public:
  92.          vector <int> baseAndExponent(string N){
  93.              int64 n = gao(N);
  94.              VI ret; ret.clear();
  95.              pii ans;
  96.              for (int i = ; i < ; ++ i){
  97.                  int tmp = gao(n, i);
  98.                  if (tmp == -) continue;
  99.                  ret.PB (tmp); ret.PB (i);
  100.              }
  101.              return ret;
  102.          }
  103.  
  104.  // BEGIN CUT HERE
  105.      public:
  106.      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(); if ((Case == -) || (Case == )) test_case_4(); if ((Case == -) || (Case == )) test_case_5(); }
  107.      //void run_test(int Case) { if ((Case == -1) || (Case == 0)) test_case_0();}
  108.      private:
  109.      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(); }
  110.      void verify_case(int Case, const vector <int> &Expected, const vector <int> &Received) { cerr << "Test Case #" << Case << "..."; if (Expected == Received) cerr << "PASSED" << endl; else { cerr << "FAILED" << endl; cerr << "\tExpected: " << print_array(Expected) << endl; cerr << "\tReceived: " << print_array(Received) << endl; } }
  111.      void test_case_0() { string Arg0 = ""; int Arr1[] = {}; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[]))); verify_case(, Arg1, baseAndExponent(Arg0)); }
  112.      void test_case_1() { string Arg0 = ""; int Arr1[] = { }; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[]))); verify_case(, Arg1, baseAndExponent(Arg0)); }
  113.      void test_case_2() { string Arg0 = ""; int Arr1[] = { }; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[]))); verify_case(, Arg1, baseAndExponent(Arg0)); }
  114.      void test_case_3() { string Arg0 = ""; int Arr1[] = { }; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[]))); verify_case(, Arg1, baseAndExponent(Arg0)); }
  115.      void test_case_4() { string Arg0 = ""; int Arr1[] = {,  }; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[]))); verify_case(, Arg1, baseAndExponent(Arg0)); }
  116.      void test_case_5() { string Arg0 = ""; int Arr1[] = {,  }; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[]))); verify_case(, Arg1, baseAndExponent(Arg0)); }
  117.  
  118.  // END CUT HERE
  119.  
  120.  };
  121.  
  122.  // BEGIN CUT HERE
  123.  int main()
  124.  {
  125.  //    freopen( "a.out" , "w" , stdout );
  126.      StrongPrimePower ___test;
  127.      ___test.run_test(-);
  128.         return ;
  129.  }
  130.  // END CUT HERE

DIV1 500pt

题意：每次翻转操作r(l, r)即是将s中(l, r)的子串反转，比如对abcde进行r(0, 3)即变为cbade。进行一系列翻转变化有一个要求，即r(l1, r1), r(l2, r2), r(l3, r3), r(l4, r4)，需l1<=l2<=l3<=l4，r4>=r3>=r2>=r1。要将string s变为string g，求最少所需翻转次数。

解法：就是一个裸DP，但是我对在字符串上的dp好像一窍不通。。。。。什么时候应该挂点字符串dp的专题来刷了。具体数组设置和状态转移方程可以看官方题解，很简单。点击打开官方题解。

tag:字符串，DP

  1.  // BEGIN CUT HERE
  2.  /*
  3.   * Author:  plum rain
  4.   * score :
  5.   */
  6.  /*
  7.  
  8.   */
  9.  // END CUT HERE
  10.  #line 11 "ReversalChain.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 CLR1(x) memset(x, -1, sizeof(x))
  39.  #define PB push_back
  40.  #define SZ(v) ((int)(v).size())
  41.  #define zero(x) (((x)>0?(x):-(x))<eps)
  42.  #define out(x) cout<<#x<<":"<<(x)<<endl
  43.  #define tst(a) cout<<#a<<endl
  44.  #define CINBEQUICKER std::ios::sync_with_stdio(false)
  45.  
  46.  typedef vector<int> VI;
  47.  typedef vector<string> VS;
  48.  typedef vector<double> VD;
  49.  typedef pair<int, int> pii;
  50.  typedef long long int64;
  51.  
  52.  const double eps = 1e-;
  53.  const double PI = atan(1.0)*;
  54.  const int maxint = ;
  55.  const int inf = maxint;
  56.  
  57.  int d[][][][];
  58.  
  59.  class ReversalChain
  60.  {
  61.      public:
  62.          int minReversal(string s, string g){
  63.              CLR (d);
  64.              int n = s.size();
  65.              for (int i = ; i < n; ++ i)
  66.                  for (int j = ; j < n; ++ j){
  67.                      if (s[i] == g[j])
  68.                          d[][i][j][] = d[][i][j][] = ;
  69.                      else
  70.                          d[][i][j][] = d[][i][j][] = inf;
  71.                  }
  72.  
  73.              for (int i = ; i <= n; ++ i)
  74.                  for (int j = ; j < n; ++ j)
  75.                      for (int k = ; k < n; ++ k){
  76.                          d[i][j][k][] = d[i][j][k][] = inf;
  77.                          if (s[j] == g[k]){
  78.                              d[i][j][k][] = min(d[i][j][k][], d[i-][j+][k+][]);
  79.                              d[i][j][k][] = min(d[i][j][k][], d[i-][j+][k+][] + );
  80.                          }
  81.                          if (s[j+i-] == g[k+i-]){
  82.                              d[i][j][k][] = min(d[i][j][k][], d[i-][j][k][]);
  83.                              d[i][j][k][] = min(d[i][j][k][], d[i-][j][k][] + );
  84.                          }
  85.                          if (s[j] == g[k+i-]){
  86.                              d[i][j][k][] = min(d[i][j][k][], d[i-][j+][k][] + );
  87.                              d[i][j][k][] = min(d[i][j][k][], d[i-][j+][k][]);
  88.                          }
  89.                          if (s[j+i-] == g[k]){
  90.                              d[i][j][k][] = min(d[i][j][k][], d[i-][j][k+][] + );
  91.                              d[i][j][k][] = min(d[i][j][k][], d[i-][j][k+][]);
  92.                          }
  93.                      }
  94.              return d[n][][][] == inf ? - : d[n][][][];
  95.          }
  96.  
  97.  // BEGIN CUT HERE
  98.      public:
  99.      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(); if ((Case == -) || (Case == )) test_case_4(); }
  100.      private:
  101.      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(); }
  102.      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; } }
  103.      void test_case_0() { string Arg0 = ""; string Arg1 = ""; int Arg2 = ; verify_case(, Arg2, minReversal(Arg0, Arg1)); }
  104.      void test_case_1() { string Arg0 = ""; string Arg1 = ""; int Arg2 = ; verify_case(, Arg2, minReversal(Arg0, Arg1)); }
  105.      void test_case_2() { string Arg0 = ""; string Arg1 = ""; int Arg2 = -; verify_case(, Arg2, minReversal(Arg0, Arg1)); }
  106.      void test_case_3() { string Arg0 = ""; string Arg1 = ""; int Arg2 = ; verify_case(, Arg2, minReversal(Arg0, Arg1)); }
  107.      void test_case_4() { string Arg0 = ""; string Arg1 = ""; int Arg2 = ; verify_case(, Arg2, minReversal(Arg0, Arg1)); }
  108.  
  109.  // END CUT HERE
  110.  
  111.  };
  112.  
  113.  // BEGIN CUT HERE
  114.  int main()
  115.  {
  116.  //    freopen( "a.out" , "w" , stdout );
  117.      ReversalChain ___test;
  118.      ___test.run_test(-);
  119.         return ;
  120.  }
  121.  // END CUT HERE

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