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SRM 400(1-250pt, 1-500pt)

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-8;
 53 const double PI = atan(1.0)*4;
 54 const int maxint = 2139062143;
 55 const int maxx = 1000005;
 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[0] = vis[1] = 1;
 65     for (int64 i = 2; i <= m; ++ i) if (!vis[i])
 66         for (int64 j = i*i; j <= n; j += i) vis[j] = 1;
 67         
 68 }
 69 
 70 int primes(int n)
 71 {
 72     sieve(n);
 73     int ret = 0;
 74     for (int i = 2; 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 <= 1000000) return (!vis[n]);
 83 
 84     for (int i = 0; i < all; ++ i) 
 85         if ((n % prm[i]) == 0){
 86             return 0;
 87         }
 88     return 1;
 89 }
 90 
 91 int gao1(int64 n)
 92 {
 93     int64 m = (int64)sqrt(n + 0.5);
 94     if (m*m != n) return -1;
 95 
 96     if (ok(m)) return m;
 97     return -1;
 98 }
 99 
100 pii gao2(int64 n)
101 {
102     int ret;
103     pii tmp;
104     for (int i = 0; i < all; ++ i) if ((n % prm[i]) == 0){
105         ret = 0;
106         while ((n % prm[i]) == 0) {
107             ++ ret;
108             n /= prm[i];
109         }
110         if (ret > 1 && n == 1){
111             tmp.first = prm[i];
112             tmp.second = ret;
113             return tmp;
114         }
115         else{
116             tmp.first = -1;
117             return tmp;
118         }
119     }
120     tmp.first = -1;
121     return tmp;
122 }
123 
124 int64 gao(string n)
125 {
126     int64 ret = 0;
127     for (int i = 0; i < n.size(); ++ i)
128         ret = ret * 10 + n[i] - '0';
129     return ret;
130 }
131 
132 class StrongPrimePower
133 {
134     public:
135         vector <int> baseAndExponent(string N){
136             all = primes(1000000);
137             int64 n = gao(N);
138             
139             int64 tmp = gao1(n);
140             vector<int> ans;
141             ans.clear();
142             if (tmp != -1){
143                 ans.PB ((int)tmp); ans.PB (2);
144                 return ans;
145             }
146             pii t = gao2(n);
147             if (t.first != -1){
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 == -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(); if ((Case == -1) || (Case == 4)) test_case_4(); if ((Case == -1) || (Case == 5)) 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 << "	Expected: " << print_array(Expected) << endl; cerr << "	Received: " << print_array(Received) << endl; } }
160     void test_case_0() { string Arg0 = "27"; int Arr1[] = {3, 3 }; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[0]))); verify_case(0, Arg1, baseAndExponent(Arg0)); }
161     void test_case_1() { string Arg0 = "10"; int Arr1[] = { }; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[0]))); verify_case(1, Arg1, baseAndExponent(Arg0)); }
162     void test_case_2() { string Arg0 = "7"; int Arr1[] = { }; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[0]))); verify_case(2, Arg1, baseAndExponent(Arg0)); }
163     void test_case_3() { string Arg0 = "1296"; int Arr1[] = { }; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[0]))); verify_case(3, Arg1, baseAndExponent(Arg0)); }
164     void test_case_4() { string Arg0 = "576460752303423488"; int Arr1[] = {2, 59 }; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[0]))); verify_case(4, Arg1, baseAndExponent(Arg0)); }
165     void test_case_5() { string Arg0 = "999999874000003969"; int Arr1[] = {999999937, 2 }; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[0]))); verify_case(5, 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(-1);
177        return 0;
178 }
179 // END CUT HERE

View Code

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-8;
 53 const double PI = atan(1.0)*4;
 54 const int maxint = 2139062143;
 55 
 56 int64 gao(string n)
 57 {
 58     int64 ret = 0;
 59     for (int i = 0; i < n.size(); ++ i)
 60         ret = ret * 10 + n[i] - '0';
 61     return ret;
 62 }
 63 
 64 int64 mypow(int64 p, int64 n)
 65 {
 66     unsigned long long ret = 1;
 67     for (int i = 0; i < n; ++ i){
 68         ret *= p;
 69         if (ret > 1e18) return -1;
 70     }
 71     return ret;
 72 }
 73 
 74 bool ok(int y)
 75 {
 76     for (int64 i = 2; i*i <= y; ++ i)
 77         if (y % i == 0) return 0;
 78     return 1;
 79 }
 80 
 81 int gao(int64 n, int num)
 82 {
 83     int x = (int)pow(n, 1.0 / (double)num), y = -1;
 84     for (int i = -1; i < 2; ++ 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 = 2; i < 61; ++ i){
 97                 int tmp = gao(n, i);
 98                 if (tmp == -1) 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 == -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(); if ((Case == -1) || (Case == 4)) test_case_4(); if ((Case == -1) || (Case == 5)) 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 << "	Expected: " << print_array(Expected) << endl; cerr << "	Received: " << print_array(Received) << endl; } }
111     void test_case_0() { string Arg0 = "639558602475808609"; int Arr1[] = {}; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[0]))); verify_case(0, Arg1, baseAndExponent(Arg0)); }
112     void test_case_1() { string Arg0 = "10"; int Arr1[] = { }; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[0]))); verify_case(1, Arg1, baseAndExponent(Arg0)); }
113     void test_case_2() { string Arg0 = "7"; int Arr1[] = { }; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[0]))); verify_case(2, Arg1, baseAndExponent(Arg0)); }
114     void test_case_3() { string Arg0 = "1296"; int Arr1[] = { }; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[0]))); verify_case(3, Arg1, baseAndExponent(Arg0)); }
115     void test_case_4() { string Arg0 = "576460752303423488"; int Arr1[] = {2, 59 }; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[0]))); verify_case(4, Arg1, baseAndExponent(Arg0)); }
116     void test_case_5() { string Arg0 = "999999874000003969"; int Arr1[] = {999999937, 2 }; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[0]))); verify_case(5, 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(-1);
128        return 0;
129 }
130 // END CUT HERE

View Code

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-8;
 53 const double PI = atan(1.0)*4;
 54 const int maxint = 2139062143;
 55 const int inf = maxint;
 56 
 57 int d[55][55][55][2];
 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 = 0; i < n; ++ i)
 66                 for (int j = 0; j < n; ++ j){
 67                     if (s[i] == g[j]) 
 68                         d[1][i][j][0] = d[1][i][j][1] = 0;
 69                     else 
 70                         d[1][i][j][0] = d[1][i][j][1] = inf;
 71                 }
 72 
 73             for (int i = 2; i <= n; ++ i)
 74                 for (int j = 0; j < n; ++ j)
 75                     for (int k = 0; k < n; ++ k){
 76                         d[i][j][k][0] = d[i][j][k][1] = inf;
 77                         if (s[j] == g[k]){
 78                             d[i][j][k][0] = min(d[i][j][k][0], d[i-1][j+1][k+1][0]);
 79                             d[i][j][k][1] = min(d[i][j][k][1], d[i-1][j+1][k+1][0] + 1);
 80                         }
 81                         if (s[j+i-1] == g[k+i-1]){
 82                             d[i][j][k][0] = min(d[i][j][k][0], d[i-1][j][k][0]);
 83                             d[i][j][k][1] = min(d[i][j][k][1], d[i-1][j][k][0] + 1);
 84                         }
 85                         if (s[j] == g[k+i-1]){
 86                             d[i][j][k][0] = min(d[i][j][k][0], d[i-1][j+1][k][1] + 1);
 87                             d[i][j][k][1] = min(d[i][j][k][1], d[i-1][j+1][k][1]);
 88                         }
 89                         if (s[j+i-1] == g[k]){
 90                             d[i][j][k][0] = min(d[i][j][k][0], d[i-1][j][k+1][1] + 1);
 91                             d[i][j][k][1] = min(d[i][j][k][1], d[i-1][j][k+1][1]);
 92                         }
 93                     }
 94             return d[n][0][0][0] == inf ? -1 : d[n][0][0][0];
 95         }
 96         
 97 // BEGIN CUT HERE
 98     public:
 99     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(); if ((Case == -1) || (Case == 4)) 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 << "	Expected: "" << Expected << '"' << endl; cerr << "	Received: "" << Received << '"' << endl; } }
103     void test_case_0() { string Arg0 = "1100"; string Arg1 = "0110"; int Arg2 = 1; verify_case(0, Arg2, minReversal(Arg0, Arg1)); }
104     void test_case_1() { string Arg0 = "111000"; string Arg1 = "101010"; int Arg2 = 2; verify_case(1, Arg2, minReversal(Arg0, Arg1)); }
105     void test_case_2() { string Arg0 = "0"; string Arg1 = "1"; int Arg2 = -1; verify_case(2, Arg2, minReversal(Arg0, Arg1)); }
106     void test_case_3() { string Arg0 = "10101"; string Arg1 = "10101"; int Arg2 = 0; verify_case(3, Arg2, minReversal(Arg0, Arg1)); }
107     void test_case_4() { string Arg0 = "111000111000"; string Arg1 = "001100110011"; int Arg2 = 4; verify_case(4, 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(-1);
119        return 0;
120 }
121 // END CUT HERE

View Code

------------------------------------------------------------------
现在的你，在干什么呢？
你是不是还记得，你说你想成为岩哥那样的人。

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原文地址：https://www.cnblogs.com/plumrain/p/srm_400.html