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  • 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

      1 // BEGIN CUT HERE
      2 /*
      3  * Author:  plum rain
      4  * score :
      5  */
      6 /*
      7 
      8  */
      9 // END CUT HERE
     10 #line 11 "OrderedSuperString.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 clr0(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 all(t) t.begin(),t.end()
     42 #define zero(x) (((x)>0?(x):-(x))<eps)
     43 #define out(x) cout<<#x<<":"<<(x)<<endl
     44 #define tst(a) cout<<a<<" "
     45 #define tst1(a) cout<<#a<<endl
     46 #define CINBEQUICKER std::ios::sync_with_stdio(false)
     47 
     48 typedef vector<int> vi;
     49 typedef vector<string> vs;
     50 typedef vector<double> vd;
     51 typedef pair<int, int> pii;
     52 typedef long long int64;
     53 
     54 const double eps = 1e-8;
     55 const double PI = atan(1.0)*4;
     56 const int inf = 2139062143 / 2;
     57 
     58 bool ok(string a, string b)
     59 {
     60     for (int i = 0; i < min(sz(a), sz(b)); ++ i)
     61         if (a[i] != b[i]) return 0;
     62     return 1;
     63 }
     64 
     65 class OrderedSuperString
     66 {
     67     public:
     68         int getLength(vector <string> w){
     69             string s; s.clear();
     70             int idx = 0;
     71             for (int i = 0; i < sz(w); ++ i){
     72                 int match = sz(s);
     73                 for (int j = idx; j < sz(s); ++ j)
     74                     if (ok(string(s.begin()+j, s.end()), w[i])){
     75                         match = j; break;
     76                     }
     77                 idx = match;
     78                 if (sz(s) - match < sz(w[i]))
     79                     s += string(w[i].begin()+sz(s)-match, w[i].end());
     80             }
     81             return sz(s);
     82         }
     83         
     84 // BEGIN CUT HERE
     85     public:
     86     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(); }
     87     private:
     88     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(); }
     89     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; } }
     90     void test_case_0() { string Arr0[] = {"aaaaaaaaaaabaaaaaaaa", "bac", "aaaabacaaa", "ab", "ba", "a", "ca"}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arg1 = 4; verify_case(0, Arg1, getLength(Arg0)); }
     91     void test_case_1() { string Arr0[] = {"a","a","b","a"}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arg1 = 3; verify_case(1, Arg1, getLength(Arg0)); }
     92     void test_case_2() { string Arr0[] = {"abcdef", "ab","bc", "de","ef"}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arg1 = 6; verify_case(2, Arg1, getLength(Arg0)); }
     93     void test_case_3() { string Arr0[] = {"ab","bc", "de","ef","abcdef"}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arg1 = 12; verify_case(3, Arg1, getLength(Arg0)); }
     94 
     95 // END CUT HERE
     96 
     97 };
     98 
     99 // BEGIN CUT HERE
    100 int main()
    101 {
    102 //    freopen( "a.out" , "w" , stdout );    
    103     OrderedSuperString ___test;
    104     ___test.run_test(-1);
    105        return 0;
    106 }
    107 // END CUT HERE
    View Code

    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

      1 // BEGIN CUT HERE
      2 /*
      3  * Author:  plum rain
      4  * score :
      5  */
      6 /*
      7 
      8  */
      9 // END CUT HERE
     10 #line 11 "MagicalSpheres.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 clr0(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 all(t) t.begin(),t.end()
     42 #define zero(x) (((x)>0?(x):-(x))<eps)
     43 #define out(x) cout<<#x<<":"<<(x)<<endl
     44 #define tst(a) cout<<a<<" "
     45 #define tst1(a) cout<<#a<<endl
     46 #define CINBEQUICKER std::ios::sync_with_stdio(false)
     47 
     48 typedef vector<int> vi;
     49 typedef vector<string> vs;
     50 typedef vector<double> vd;
     51 typedef pair<int, int> pii;
     52 typedef long long int64;
     53 
     54 const double eps = 1e-8;
     55 const double PI = atan(1.0)*4;
     56 const int inf = 2139062143 / 2;
     57 const int N = 100005;
     58 
     59 
     60 class MagicalSpheres
     61 {
     62     public:
     63         int64 gao (int x, int m)
     64         {
     65             int64 num = 0;
     66             int64 tmp = m;
     67             while (tmp <= x){
     68                 num += x / tmp;
     69                 tmp *= m;
     70             }
     71             return num;
     72         }
     73         
     74         int divideWork(int n1, int n2, int up){
     75             for (int i = up; i; -- i){
     76                 bool ok = 0;
     77                 int k = i;
     78                 for (int j = 2; j*j <= k; ++ j) if (k % j == 0){
     79                     int t = 0;
     80                     while (!(k % j))
     81                         k /= j, ++ t;
     82                     if (gao(n1+n2, j) - gao(n1, j) - gao(n2, j) < t) ok = 1;
     83                 }
     84                 if (k != 1)
     85                     if (gao(n1+n2, k) - gao(n1, k) - gao(n2, k) < 1) ok = 1;
     86                 if (!ok) return i;
     87             }
     88             return 1;
     89         }
     90 
     91         // BEGIN CUT HERE
     92     public:
     93         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(); }
     94     private:
     95         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(); }
     96         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; } }
     97         void test_case_0() { int Arg0 = 1000000000; int Arg1 = 1000000000; int Arg2 = 100000; int Arg3 = 2; verify_case(0, Arg3, divideWork(Arg0, Arg1, Arg2)); }
     98         void test_case_1() { int Arg0 = 3; int Arg1 = 3; int Arg2 = 50; int Arg3 = 20; verify_case(1, Arg3, divideWork(Arg0, Arg1, Arg2)); }
     99         void test_case_2() { int Arg0 = 4; int Arg1 = 3; int Arg2 = 4; int Arg3 = 1; verify_case(2, Arg3, divideWork(Arg0, Arg1, Arg2)); }
    100         void test_case_3() { int Arg0 = 15634; int Arg1 = 456; int Arg2 = 5000; int Arg3 = 4990; verify_case(3, Arg3, divideWork(Arg0, Arg1, Arg2)); }
    101 
    102         // END CUT HERE
    103 
    104 };
    105 
    106 // BEGIN CUT HERE
    107 int main()
    108 {
    109 //    freopen( "a.out" , "w" , stdout );    
    110     MagicalSpheres ___test;
    111     ___test.run_test(-1);
    112        return 0;
    113 }
    114 // END CUT HERE
    View Code
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  • 原文地址:https://www.cnblogs.com/plumrain/p/srm_409.html
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