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 << "\tExpected: \"" << Expected << '\"' << endl; cerr << "\tReceived: \"" << 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 << "\tExpected: \"" << Expected << '\"' << endl; cerr << "\tReceived: \"" << 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

 

 

 

 

 

posted @ 2013-12-28 23:01  Plumrain  阅读(267)  评论(0)  编辑  收藏  举报