SRM 358(1-250,500pt)

DIV1 250pt

题意：电视目前停留在第100台，有一个遥控器，可以向上或向下换台(需要按键一次)，也可以按一些数字，然后直接跳到该台(需要按键次数等于数字数，不需要按确定键)。但是，这个遥控一些数字键是坏的不能按。问要换到x台最少需要按多少次。x <= 500000。

解法：直接搜索。可能用bfs会快点，但我更喜欢写dfs就用了。

tag:search

 // BEGIN CUT HERE
 /*

 */
 // END CUT HERE
 #line 7 "BrokenButtons.cpp"
 #include <cstdlib>
 #include <cctype>
 #include <cstring>
 #include <cstdio>
 #include <cmath>
 #include <algorithm>
 #include <vector>
 #include <iostream>
 #include <sstream>
 #include <set>
 #include <queue>
 #include <fstream>
 #include <numeric>
 #include <iomanip>
 #include <bitset>
 #include <list>
 #include <stdexcept>
 #include <functional>
 #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 mp make_pair
 #define sz(v) ((int)(v).size())
 #define out(x) cout<<#x<<":"<<(x)<<endl
 #define tst(x) cout<<x<<":"<<" "
 #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 long long int64;

 const double eps = 1e-;
 const double PI = atan(1.0)*;
 const int inf =  / ;

 inline int MyMod( int a , int b ) { return (a%b+b)%b;}

 class BrokenButtons
 {
     public:
         int ans, len, p;
         bool v[];
         void dfs(int x, int num)
         {
             if (num) ans = min(ans, abs(p-x) + num);
             if (num <= len)
                 for (int i = ; i < ; ++ i) if (!v[i]) dfs (x*+i, num+);
         }
         int minPresses(int pag, vector <int> bro){
             clr0 (v);
             for (int i = ; i < sz(bro); ++ i) v[bro[i]] = ;
             ans = abs(pag-); p = pag; len = ;
             while (pag) pag /= , len ++;
             dfs (, );
             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 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; } }
     void test_case_0() { int Arg0 = ; int Arr1[] = { , ,  }; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[]))); int Arg2 = ; verify_case(, Arg2, minPresses(Arg0, Arg1)); }
     void test_case_1() { int Arg0 = ; int Arr1[] = { , ,  }; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[]))); int Arg2 = ; verify_case(, Arg2, minPresses(Arg0, Arg1)); }
     void test_case_2() { int Arg0 = ; int Arr1[] = { }; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[]))); int Arg2 = ; verify_case(, Arg2, minPresses(Arg0, Arg1)); }
     void test_case_3() { int Arg0 = ; int Arr1[] = {, , , , , , , , }; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[]))); int Arg2 = ; verify_case(, Arg2, minPresses(Arg0, Arg1)); }
     void test_case_4() { int Arg0 = ; int Arr1[] = { ,  }; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[]))); int Arg2 = ; verify_case(, Arg2, minPresses(Arg0, Arg1)); }

 // END CUT HERE

 };
 //by plum rain
 // BEGIN CUT HERE
 int main()
 {
     //freopen( "a.out" , "w" , stdout );
     BrokenButtons ___test;
     ___test.run_test(-);
        return ;
 }
 // END CUT HERE

DIV1 500pt

题意：有一个数字a[]，a的某子数组b[]如果能表示出a，则称b为好子数组。能表示出的意思即为，可以写出如下等式：对任意a[i]均有，a[i] = b[0]*t0 + b[1]*t1 + b[2]*t2 + b[3]*t3 + ... + b[m-1]*t(m-1).(b中有m个元素)。给定a，求它的元素数量最少的好子数组b。

　　　a[i] <= 10^7

解法：首先，从a中选出一些元素形成所求的b，其余元素构成数组b'。则一定有，b中所有元素的最大公约数一定是b'中所有元素的最大公约数的约数，否则无法表示。

　　　所以，为了方便，先将a中所有元素都除以最大公约数，然后在其中找到最少个数，使得找出的数最大公约数为1即可。

　　　其次，注意到，a[i]中每个元素含有质因子的个数最多为8个。(2*3*...23>10^7)

　　　所以，由于b中至少含有一个元素，所以对所有a[i]，枚举a[i]在b中的时候，b中最少含有多少个数，问题可以由集合dp得到解决。

tag:math, dp, good

 // BEGIN CUT HERE
 /*

 */
 // END CUT HERE
 #line 7 "BalanceScale.cpp"
 #include <cstdlib>
 #include <cctype>
 #include <cstring>
 #include <cstdio>
 #include <cmath>
 #include <algorithm>
 #include <vector>
 #include <iostream>
 #include <sstream>
 #include <set>
 #include <queue>
 #include <fstream>
 #include <numeric>
 #include <iomanip>
 #include <bitset>
 #include <list>
 #include <stdexcept>
 #include <functional>
 #include <string>
 #include <utility>
 #include <map>
 #include <ctime>
 #include <stack>

 using namespace std;

 #define clr(x) memset(x, 0, sizeof(x))
 #define clrs(x,y) memset(x, y, sizeof(x))
 #define pb push_back
 #define sz(v) ((int)(v).size())
 #define all(v) (v).begin(), (v).end()
 #define out(x) cout<<#x<<":"<<(x)<<endl
 #define tst(x) cout<<x<<" "
 #define tst1(a) cout<<#a<<endl
 #define CINBEQUICKER std::ios::sync_with_stdio(false)

 typedef long long int64;
 typedef vector<int> vi;
 typedef vector<string> vs;
 typedef vector<double> vd;
 typedef pair<int, int> pii;

 const double eps = 1e-;
 const double PI = atan(1.0)*;
 const int inf =  / ;

 inline int MyMod( int a , int b ) { return (a%b+b)%b;}

 int d[][<<];
 vi divv;

 class BalanceScale
 {
     public:
         int gao(int sta, int x)
         {
             int m = sz(divv), ret = ;
             for (int i = ; i < m; ++ i) if (sta & (<<i)){
                 if (x % divv[i] == ) ret |=  << i;
             }
             return ret;
         }
         int takeWeights(vector <int> w){
             sort(all(w));
             int gd = w[], n = sz(w);
             for (int i = ; i < n; ++ i) gd = __gcd(gd, w[i]);
             for (int i = ; i < n; ++ i){
                 w[i] /= gd;
                 if (w[i] == ) return ;
             }
             int ans = n;
             for (int i = ; i < n; ++ i){
                 int tmp = w[i];
                 divv.clear();
                 for (int64 j = ; j*j <= tmp; ++ j) if (tmp % j == ){
                     divv.pb (j);
                     while (tmp % j == ) tmp /= j;
                 }
                 if (tmp != ) divv.pb (tmp);

                 int m = sz(divv);
                 clr (d); d[][(<<m)-] = ;
                 for (int j = ; j < n; ++ j){
                     for (int k = ; k <= j; ++ k)
                         for (int t = ; t < (<<m); ++ t)
                             if (d[k][t]) d[k+][gao(t, w[j])] = ;
                 }

                 for (int j = ; j < n; ++ j) if (d[j][]) ans = min(ans, j+);
             }
             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(); }
     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 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; } }
     //void test_case_0() { int Arr0[] = { 5, 4, 1, 8 }; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arg1 = 1; verify_case(0, Arg1, takeWeights(Arg0)); }
     void test_case_0() { int Arr0[] = {, , , , }; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arg1 = ; verify_case(, Arg1, takeWeights(Arg0)); }
     void test_case_1() { int Arr0[] = { , , ,  }; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arg1 = ; verify_case(, Arg1, takeWeights(Arg0)); }
     void test_case_2() { int Arr0[] = { , , ,  }; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arg1 = ; verify_case(, Arg1, takeWeights(Arg0)); }
     void test_case_3() { int Arr0[] = { , ,  }; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[]))); int Arg1 = ; verify_case(, Arg1, takeWeights(Arg0)); }

 // END CUT HERE

 };
 //by plum rain
 // BEGIN CUT HERE
 int main()
 {
     //freopen( "a.out" , "w" , stdout );
     BalanceScale ___test;
     ___test.run_test(-);
        return ;
 }
 // END CUT HERE