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

  1 // BEGIN CUT HERE
  2 /*
  3  * Author:  plum rain
  4  * score :
  5  */
  6 /*
  7 
  8  */
  9 // END CUT HERE
 10 #line 11 "ConcatenateNumber.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 class ConcatenateNumber
 59 {
 60     public:
 61         bool an[100005];
 62         int getSmallest(int n, int k){
 63             clr0 (an);
 64             int64 nn = n, tmp = n % k, ten = 1;
 65             while (n)
 66                 ten *= 10, n /= 10;
 67             int cnt = 1;
 68             while (1){
 69                 if (tmp == 0) return cnt;
 70                 if (an[tmp]) return -1;
 71                 else an[tmp] = 1;
 72 
 73                 tmp = (tmp * ten + nn)% k;
 74                 ++ cnt;
 75             }
 76         }
 77         
 78 // BEGIN CUT HERE
 79     public:
 80     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(); if ((Case == -1) || (Case == 4)) test_case_4(); }
 81     private:
 82     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(); }
 83     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; } }
 84     void test_case_0() { int Arg0 = 2; int Arg1 = 9; int Arg2 = 9; verify_case(0, Arg2, getSmallest(Arg0, Arg1)); }
 85     void test_case_1() { int Arg0 = 121; int Arg1 = 11; int Arg2 = 1; verify_case(1, Arg2, getSmallest(Arg0, Arg1)); }
 86     void test_case_2() { int Arg0 = 1; int Arg1 = 2; int Arg2 = -1; verify_case(2, Arg2, getSmallest(Arg0, Arg1)); }
 87     void test_case_3() { int Arg0 = 35; int Arg1 = 98765; int Arg2 = 9876; verify_case(3, Arg2, getSmallest(Arg0, Arg1)); }
 88     void test_case_4() { int Arg0 = 1000000000; int Arg1 = 3; int Arg2 = 3; verify_case(4, Arg2, getSmallest(Arg0, Arg1)); }
 89 
 90 // END CUT HERE
 91 
 92 };
 93 
 94 // BEGIN CUT HERE
 95 int main()
 96 {
 97 //    freopen( "a.out" , "w" , stdout );    
 98     ConcatenateNumber ___test;
 99     ___test.run_test(-1);
100        return 0;
101 }
102 // END CUT HERE
View Code

 

posted @ 2013-12-28 02:36  Plumrain  阅读(141)  评论(0)  编辑  收藏  举报