SRM 405(1-250pt, 1-500pt)

DIV1 250pt

题意：以linux系统中文件系统的路径表示方法为背景，告诉你某文件的绝对路径和当前位置，求相对路径。具体看样例。

解法：模拟题，不多说。每次碰到STL的题自己的代码都会显得很sb...同时速度真的太不够了....

tag:模拟题

 // BEGIN CUT HERE

 /*

  * Author:  plum rain

  * score :

  */

 /*

  */

 // END CUT HERE

 #line 11 "RelativePath.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 ALL(t) t.begin(),t.end()

 #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 = ;

 VS p, c;

 string temp = "../";

 class RelativePath

 {

     public:

         string makeRelative(string pat, string cur){

             p.clear(); c.clear();

             string s;

             for (int i = ; i < pat.size(); ++ i){

                 s.clear();

                 while (i < (int)pat.size() && pat[i] != '/'){

                     s.PB (pat[i]); ++ i;

                 }

                 if (s.size()) p.PB (s);

             }

             for (int i = ; i < cur.size(); ++ i){

                 s.clear();

                 while (i < cur.size() && cur[i] != '/'){

                     s.PB (cur[i]); ++ i;

                 }

                 if (s.size()) c.PB (s);

             }

             int t1 = , t2 = ;

             while (t1 < p.size() && t2 < c.size()){

                 if (p[t1] == c[t2])

                     ++ t1, ++ t2;

                 else break;

             }

             string ans; ans.clear();

             for (int i = ; i < c.size()-t2; ++ i) ans += temp;

             for (int i = t1; i < p.size(); ++ i){

                 if (i != t1) ans.PB ('/');

                 ans += p[i];

             }

             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(); }

     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 string &Expected, const string &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 = "/home/top/data/file"; string Arg1 = "/home/user/pictures"; string Arg2 = "../../top/data/file"; verify_case(, Arg2, makeRelative(Arg0, Arg1)); }

     void test_case_1() { string Arg0 = "/home/user/movies/title"; string Arg1 = "/home/user/pictures"; string Arg2 = "../movies/title"; verify_case(, Arg2, makeRelative(Arg0, Arg1)); }

     void test_case_2() { string Arg0 = "/file"; string Arg1 = "/"; string Arg2 = "file"; verify_case(, Arg2, makeRelative(Arg0, Arg1)); }

     void test_case_3() { string Arg0 = "/a/b/a/b/a/b"; string Arg1 = "/a/b/a/a/b/a/b"; string Arg2 = "../../../../b/a/b"; verify_case(, Arg2, makeRelative(Arg0, Arg1)); }

     void test_case_4() { string Arg0 = "/root/root/root"; string Arg1 = "/root"; string Arg2 = "root/root"; verify_case(, Arg2, makeRelative(Arg0, Arg1)); }

 // END CUT HERE

 };

 // BEGIN CUT HERE

 int main()

 {

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

     RelativePath ___test;

     ___test.run_test(-);

        return ;

 }

 // END CUT HERE

DIV1 500pt

题意：对于一个边权均为1的强连通图，每两点之间有且仅有一条路。若能从任意点走x步之后回到该点，则称x为幸运数。用一个数列a[](0-based)表示是否是幸运数，如果i为幸运数a[i-1] = 1，否则a[i-1] = 0。可以证明，a[]一定会出现循环，找出最小的循环节和它最早出现的位置。比如“0011011111111”可以表示成"00110111(11)"，也可以表示成 "00110(1)"，但是要返回后者。n <= 30。

解法：首先，给出两个结论：1、若x为幸运数，则x一定是a[]的循环节之一；2、若x，y均为a[]的循环节，则gcd(x, y)也为a[]的循环节；3、若x，y均为循环节，则第一个循环节出现的位置不会在x*y之后。

　　　那么，这道题的数据范围就被限制出来了。循环节最长为30，返回的答案长度最长为900 + 30 + 2。所以，直接暴力2000以内所有的数是否是幸运数，然后再暴力枚举循环节长度和循环起始位置即可。。。。。。

tag：think, good

 // BEGIN CUT HERE

 /*

  * Author:  plum rain

  * score :

  */

 /*

  */

 // END CUT HERE

 #line 11 "AllCycleLengths.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 clr0(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 all(t) t.begin(),t.end()

 #define zero(x) (((x)>0?(x):-(x))<eps)

 #define out(x) cout<<#x<<":"<<(x)<<endl

 #define tst(a) cout<<a<<" "

 #define tst1(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 map<int, int> mpii;

 typedef long long int64;

 const double eps = 1e-;

 const double PI = atan(1.0)*;

 const int inf =  / ;

 const int N = ;

 int n;

 int con[], tcon[];

 class AllCycleLengths

 {

     public:

         string findAll(vector <string> arc){

             n = sz(arc);

             clr0 (con);

             set<int> st;

             VI a, b;

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

                 b.clear(); a.clear(); a.pb (i);

                 int times = ;

                 while (times <= N){

                     ++ times;

                     set<int> st;

                     for (int j = ; j < sz(a); ++ j)

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

                             if (arc[a[j]][k] == 'Y' && !st.count(k)){

                                 b.pb (k); st.insert(k);

                             }

                     for (int j = ; j < sz(b); ++ j)

                         if (b[j] == i) con[times] = ;

                     a = b; b.clear();

                 }

             }

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

                 int pos = -, end = -;

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

                     int ok = , flag = i + rec, times = ;

                     while (times <  && flag + rec < N){

                         ok = ;

                         for (int t = ; t < rec; ++ t)

                             if (con[i+t] != con[flag+t]){

                                 ok = ; break;

                             }

                         if (!ok) break;

                         ++ times; flag += rec;

                     }

                     if (ok){

                         pos = i; end = pos + rec - ;

                         break;

                     }

                 }

                 if (pos == -) continue;

                 out (pos);

                 string ret; ret.clear();

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

                     if (i == pos) ret.pb ('(');

                     ret.pb (con[i] + '');

                 }

                 ret.pb (')');

                 return ret;

             }

             return "(0)";

         }

 // 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(); }

     //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 string &Expected, const string &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 Arr0[] = {"NNNNNNNNNNNNNNNNNNNNNNYNN", "NNNNNNNNNNNYNNNNNNNNNNNNN", "NNNNNNNNNNNNYNNNNNNNNNNNN", "NNNNNNNYNNNNNNNNNNNNNNNNN", "NNNNNNNNNNNNNYNNNNNNNNNNN", "NNNNNNYNNNNNNNNNNNNNNNNNN", "NNNNNNNNNNNNNNNNNNNNNYNNN", "NNNNNNNNNNNNNNNNNNNNNNNNY", "NNYNNNNNNNNNNNYNNNNNNNNNN", "NNNNNNNNNNNNNNNYNNNNNNNNN", "YNNNNNNNNNNNNNNNNNNNNNNNN", "NNNNYNNNNNNNNNNNNNNNNNNNN", "NNNNNNNNNNNNNNNNNNNNNNNYN", "NNNNNNNNYNNNNNNNNNNNNNNNN", "NNYNNNNNNNNNNNNNNNNNNNNNN", "NNNNNNNNNNYNNNNNNNNNNNNNN", "NNNNNNNNNNNNNNNNNNNNYNNNN", "NNNYNNNNNNNNNNNNNNNNNNNNN", "NNNNNNNNNYNNNNNNNNNNNNNNN", "NNNNNNNNNNNNNNNNYNNNNNNNN", "NYNNNNNNNNNNNNNNNNNNNNNNN", "NNNNNNNNNNNNNNNNNYNNNNNNN", "NNNNNNNNNNNNNNNNNNNYNNNNN", "NNNNNYNNNNNNNNNNNNNNNNNNN", "NNNNNNNNNNNNNNNNNNYNNNNNN"}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); string Arg1 = "0010010110110(1)"; verify_case(, Arg1, findAll(Arg0)); }

     //void test_case_0() { string Arr0[] = {"NYNN", "NNYY", "NNNY", "YNNN"}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); string Arg1 = "00110(1)"; verify_case(0, Arg1, findAll(Arg0)); }

     void test_case_1() { string Arr0[] = {"NY", "YN"}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); string Arg1 = "(01)"; verify_case(, Arg1, findAll(Arg0)); }

     void test_case_2() { string Arr0[] = {"NYYYY", "NNYYY", "NNNYY", "NNNNY", "YNNNN"}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); string Arg1 = "0(1)"; verify_case(, Arg1, findAll(Arg0)); }

     void test_case_3() { string Arr0[] = {"NYNNN", "NNYNN", "NNNYN", "NNNNY", "YNNYN"}; vector <string> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); string Arg1 = "010(1)"; verify_case(, Arg1, findAll(Arg0)); }

 // END CUT HERE

 };

 // BEGIN CUT HERE

 int main()

 {

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

     AllCycleLengths ___test;

     ___test.run_test(-);

        return ;

 }

 // END CUT HERE

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