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:
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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
解法二:
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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
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
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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