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：

 // BEGIN CUT HERE

 /*

  * Author:  plum rain

  * score :

  */

 /*

  */

 // END CUT HERE

 #line 11 "StrongPrimePower.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 CLR1(x) memset(x, -1, 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 pair<int, int> pii;

 typedef long long int64;

 const double eps = 1e-;

 const double PI = atan(1.0)*;

 const int maxint = ;

 const int maxx = ;

 int all;

 bool vis[maxx];

 int prm[maxx];

 void sieve(int n)

 {

     int m = (int)sqrt(n+0.5);

     CLR (vis); vis[] = vis[] = ;

     for (int64 i = ; i <= m; ++ i) if (!vis[i])

         for (int64 j = i*i; j <= n; j += i) vis[j] = ;

 }

 int primes(int n)

 {

     sieve(n);

     int ret = ;

     for (int i = ; i <= n; ++ i)

         if (!vis[i]) prm[ret++] = i;

     return ret;

 }

 bool ok(int64 n)

 {

     if (n <= ) return (!vis[n]);

     for (int i = ; i < all; ++ i)

         if ((n % prm[i]) == ){

             return ;

         }

     return ;

 }

 int gao1(int64 n)

 {

     int64 m = (int64)sqrt(n + 0.5);

     if (m*m != n) return -;

     if (ok(m)) return m;

     return -;

 }

 pii gao2(int64 n)

 {

     int ret;

     pii tmp;

     for (int i = ; i < all; ++ i) if ((n % prm[i]) == ){

         ret = ;

         while ((n % prm[i]) == ) {

             ++ ret;

             n /= prm[i];

         }

         if (ret >  && n == ){

             tmp.first = prm[i];

             tmp.second = ret;

             return tmp;

         }

         else{

             tmp.first = -;

             return tmp;

         }

     }

     tmp.first = -;

     return tmp;

 }

 int64 gao(string n)

 {

     int64 ret = ;

     for (int i = ; i < n.size(); ++ i)

         ret = ret *  + n[i] - '';

     return ret;

 }

 class StrongPrimePower

 {

     public:

         vector <int> baseAndExponent(string N){

             all = primes();

             int64 n = gao(N);

             int64 tmp = gao1(n);

             vector<int> ans;

             ans.clear();

             if (tmp != -){

                 ans.PB ((int)tmp); ans.PB ();

                 return ans;

             }

             pii t = gao2(n);

             if (t.first != -){

                 ans.PB (t.first); ans.PB (t.second);

                 return ans;

             }

             return ans;

         }

 // 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(); if ((Case == -) || (Case == )) test_case_4(); if ((Case == -) || (Case == )) test_case_5(); }

     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 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; } }

     void test_case_0() { string Arg0 = ""; int Arr1[] = {,  }; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[]))); verify_case(, Arg1, baseAndExponent(Arg0)); }

     void test_case_1() { string Arg0 = ""; int Arr1[] = { }; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[]))); verify_case(, Arg1, baseAndExponent(Arg0)); }

     void test_case_2() { string Arg0 = ""; int Arr1[] = { }; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[]))); verify_case(, Arg1, baseAndExponent(Arg0)); }

     void test_case_3() { string Arg0 = ""; int Arr1[] = { }; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[]))); verify_case(, Arg1, baseAndExponent(Arg0)); }

     void test_case_4() { string Arg0 = ""; int Arr1[] = {,  }; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[]))); verify_case(, Arg1, baseAndExponent(Arg0)); }

     void test_case_5() { string Arg0 = ""; int Arr1[] = {,  }; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[]))); verify_case(, Arg1, baseAndExponent(Arg0)); }

 // END CUT HERE

 };

 // BEGIN CUT HERE

 int main()

 {

 //    freopen( "a.out" , "w" , stdout );

     StrongPrimePower ___test;

     ___test.run_test(-);

        return ;

 }

 // END CUT HERE

解法二：

 // BEGIN CUT HERE

 /*

  * Author:  plum rain

  * score :

  */

 /*

  */

 // END CUT HERE

 #line 11 "StrongPrimePower.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 CLR1(x) memset(x, -1, 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 pair<int, int> pii;

 typedef long long int64;

 const double eps = 1e-;

 const double PI = atan(1.0)*;

 const int maxint = ;

 int64 gao(string n)

 {

     int64 ret = ;

     for (int i = ; i < n.size(); ++ i)

         ret = ret *  + n[i] - '';

     return ret;

 }

 int64 mypow(int64 p, int64 n)

 {

     unsigned long long ret = ;

     for (int i = ; i < n; ++ i){

         ret *= p;

         if (ret > 1e18) return -;

     }

     return ret;

 }

 bool ok(int y)

 {

     for (int64 i = ; i*i <= y; ++ i)

         if (y % i == ) return ;

     return ;

 }

 int gao(int64 n, int num)

 {

     int x = (int)pow(n, 1.0 / (double)num), y = -;

     for (int i = -; i < ; ++ i)

         if (mypow(x+i,num) == n && ok(x+i)) y = x + i;

     return y;

 }

 class StrongPrimePower

 {

     public:

         vector <int> baseAndExponent(string N){

             int64 n = gao(N);

             VI ret; ret.clear();

             pii ans;

             for (int i = ; i < ; ++ i){

                 int tmp = gao(n, i);

                 if (tmp == -) continue;

                 ret.PB (tmp); ret.PB (i);

             }

             return ret;

         }

 // 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(); if ((Case == -) || (Case == )) test_case_4(); if ((Case == -) || (Case == )) test_case_5(); }

     //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 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; } }

     void test_case_0() { string Arg0 = ""; int Arr1[] = {}; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[]))); verify_case(, Arg1, baseAndExponent(Arg0)); }

     void test_case_1() { string Arg0 = ""; int Arr1[] = { }; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[]))); verify_case(, Arg1, baseAndExponent(Arg0)); }

     void test_case_2() { string Arg0 = ""; int Arr1[] = { }; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[]))); verify_case(, Arg1, baseAndExponent(Arg0)); }

     void test_case_3() { string Arg0 = ""; int Arr1[] = { }; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[]))); verify_case(, Arg1, baseAndExponent(Arg0)); }

     void test_case_4() { string Arg0 = ""; int Arr1[] = {,  }; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[]))); verify_case(, Arg1, baseAndExponent(Arg0)); }

     void test_case_5() { string Arg0 = ""; int Arr1[] = {,  }; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[]))); verify_case(, Arg1, baseAndExponent(Arg0)); }

 // END CUT HERE

 };

 // BEGIN CUT HERE

 int main()

 {

 //    freopen( "a.out" , "w" , stdout );

     StrongPrimePower ___test;

     ___test.run_test(-);

        return ;

 }

 // 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

 // BEGIN CUT HERE

 /*

  * Author:  plum rain

  * score :

  */

 /*

  */

 // END CUT HERE

 #line 11 "ReversalChain.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 CLR1(x) memset(x, -1, 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 pair<int, int> pii;

 typedef long long int64;

 const double eps = 1e-;

 const double PI = atan(1.0)*;

 const int maxint = ;

 const int inf = maxint;

 int d[][][][];

 class ReversalChain

 {

     public:

         int minReversal(string s, string g){

             CLR (d);

             int n = s.size();

             for (int i = ; i < n; ++ i)

                 for (int j = ; j < n; ++ j){

                     if (s[i] == g[j])

                         d[][i][j][] = d[][i][j][] = ;

                     else

                         d[][i][j][] = d[][i][j][] = inf;

                 }

             for (int i = ; i <= n; ++ i)

                 for (int j = ; j < n; ++ j)

                     for (int k = ; k < n; ++ k){

                         d[i][j][k][] = d[i][j][k][] = inf;

                         if (s[j] == g[k]){

                             d[i][j][k][] = min(d[i][j][k][], d[i-][j+][k+][]);

                             d[i][j][k][] = min(d[i][j][k][], d[i-][j+][k+][] + );

                         }

                         if (s[j+i-] == g[k+i-]){

                             d[i][j][k][] = min(d[i][j][k][], d[i-][j][k][]);

                             d[i][j][k][] = min(d[i][j][k][], d[i-][j][k][] + );

                         }

                         if (s[j] == g[k+i-]){

                             d[i][j][k][] = min(d[i][j][k][], d[i-][j+][k][] + );

                             d[i][j][k][] = min(d[i][j][k][], d[i-][j+][k][]);

                         }

                         if (s[j+i-] == g[k]){

                             d[i][j][k][] = min(d[i][j][k][], d[i-][j][k+][] + );

                             d[i][j][k][] = min(d[i][j][k][], d[i-][j][k+][]);

                         }

                     }

             return d[n][][][] == inf ? - : d[n][][][];

         }

 // 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(); if ((Case == -) || (Case == )) test_case_4(); }

     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 Arg0 = ""; string Arg1 = ""; int Arg2 = ; verify_case(, Arg2, minReversal(Arg0, Arg1)); }

     void test_case_1() { string Arg0 = ""; string Arg1 = ""; int Arg2 = ; verify_case(, Arg2, minReversal(Arg0, Arg1)); }

     void test_case_2() { string Arg0 = ""; string Arg1 = ""; int Arg2 = -; verify_case(, Arg2, minReversal(Arg0, Arg1)); }

     void test_case_3() { string Arg0 = ""; string Arg1 = ""; int Arg2 = ; verify_case(, Arg2, minReversal(Arg0, Arg1)); }

     void test_case_4() { string Arg0 = ""; string Arg1 = ""; int Arg2 = ; verify_case(, Arg2, minReversal(Arg0, Arg1)); }

 // END CUT HERE

 };

 // BEGIN CUT HERE

 int main()

 {

 //    freopen( "a.out" , "w" , stdout );

     ReversalChain ___test;

     ___test.run_test(-);

        return ;

 }

 // END CUT HERE

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