SRM 390(1-250pt)

DIV1 250pt

题意：给定整数n和k，问最少需要多少个n连接在一起形成的新整数t，使得t是k的倍数。如果不能形成倍数，输出-1。k <= 10^5，n <= 10^9。

解法：不断增加连接的n的数量，如果新形成的数除以k的余数已经出现过，说明出现循环，说明该输出-1。否则，最多执行k次就能得到答案。所以，总的来说最多执行k次，可以直接暴力。

tag:brute-force

 // BEGIN CUT HERE
 /*
  * Author:  plum rain
  * score :
  */
 /*

  */
 // END CUT HERE
 #line 11 "ConcatenateNumber.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 long long int64;

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

 class ConcatenateNumber
 {
     public:
         bool an[];
         int getSmallest(int n, int k){
             clr0 (an);
             int64 nn = n, tmp = n % k, ten = ;
             while (n)
                 ten *= , n /= ;
             int cnt = ;
             while (){
                 if (tmp == ) return cnt;
                 if (an[tmp]) return -;
                 else an[tmp] = ;

                 tmp = (tmp * ten + nn)% k;
                 ++ cnt;
             }
         }

 // 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 Arg1 = ; int Arg2 = ; verify_case(, Arg2, getSmallest(Arg0, Arg1)); }
     void test_case_1() { int Arg0 = ; int Arg1 = ; int Arg2 = ; verify_case(, Arg2, getSmallest(Arg0, Arg1)); }
     void test_case_2() { int Arg0 = ; int Arg1 = ; int Arg2 = -; verify_case(, Arg2, getSmallest(Arg0, Arg1)); }
     void test_case_3() { int Arg0 = ; int Arg1 = ; int Arg2 = ; verify_case(, Arg2, getSmallest(Arg0, Arg1)); }
     void test_case_4() { int Arg0 = ; int Arg1 = ; int Arg2 = ; verify_case(, Arg2, getSmallest(Arg0, Arg1)); }

 // END CUT HERE

 };

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