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-8;
const double PI = atan(1.0)*4;
const int inf = 2139062143 / 2;

class ConcatenateNumber
{
public:
bool an[100005];
int getSmallest(int n, int k){
clr0 (an);
int64 nn = n, tmp = n % k, ten = 1;
while (n)
ten *= 10, n /= 10;
int cnt = 1;
while (1){
if (tmp == 0) return cnt;
if (an[tmp]) return -1;
else an[tmp] = 1;

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

// BEGIN CUT HERE
public:
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(); }
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 = 2; int Arg1 = 9; int Arg2 = 9; verify_case(0, Arg2, getSmallest(Arg0, Arg1)); }
void test_case_1() { int Arg0 = 121; int Arg1 = 11; int Arg2 = 1; verify_case(1, Arg2, getSmallest(Arg0, Arg1)); }
void test_case_2() { int Arg0 = 1; int Arg1 = 2; int Arg2 = -1; verify_case(2, Arg2, getSmallest(Arg0, Arg1)); }
void test_case_3() { int Arg0 = 35; int Arg1 = 98765; int Arg2 = 9876; verify_case(3, Arg2, getSmallest(Arg0, Arg1)); }
void test_case_4() { int Arg0 = 1000000000; int Arg1 = 3; int Arg2 = 3; verify_case(4, Arg2, getSmallest(Arg0, Arg1)); }

// END CUT HERE

};

// BEGIN CUT HERE
int main()
{
//    freopen( "a.out" , "w" , stdout );
ConcatenateNumber ___test;
___test.run_test(-1);
return 0;
}
// END CUT HERE

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