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DivisiblePermutations.cpp
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DivisiblePermutations.cpp
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#include <vector>
#include <list>
#include <map>
#include <set>
#include <queue>
#include <deque>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <cassert>
#include <climits>
#include <cstring>
using namespace std;
typedef long long int64;
typedef vector<int> vi;
typedef string ST;
typedef stringstream SS;
typedef vector< vector <int> > vvi;
typedef pair<int,int> ii;
typedef vector <string> vs;
#define endl ("\n")
#define DEBUG(x) cout << #x << " = " << x << "\n"
#define Pf printf
#define Sf scanf
#define ep 1e-9
#define PI M_PI
#define E M_E
#define CL(a, b) memset(a, b, sizeof(a))
#define mp make_pair
#define pb push_back
#define all(c) (c).begin(), (c).end()
#define tr(i, c) for(__typeof((c).begin()) i = (c).begin(); i != (c).end(); i++)
#define present(x, c) ((c).find(x) != (c).end()) //map & set//
#define cpresent(x, c) (find(all(c),x) != (c).end()) //vector & list//
#define forn(i, n) for(int i = 0, loop_ends_here = (int)n; i < loop_ends_here ; i++)
#define forab(i, a, b) for(int i = a, loop_ends_here = (int)b; i <= loop_ends_here; i++)
#define rep(i, a, b) for(int i = a, loop_ends_here = (int)b; i >= loop_ends_here; i--)
#define read(n) scanf("%d", &n)
#define write(n) printf("%d ", n)
#define writeln(n) printf("%d\n", n)
class DivisiblePermutations
{
public:
long long count(string N, int M);
};
// N.size() <= 15, M <= 50
int64 fact(int n) { return n == 0 ? 1LL : n*fact(n-1); }
ST N;
int sz;
int M ;
int64 Ans[1<<16][55];
int64 solve(int mask, int rem) {
if(mask == 0)
return rem == 0 ? 1 : 0;
int64 &ret = Ans[mask][rem];
if(ret > -1)
return ret;
ret = 0;
forn(i, sz) if(mask & (1<<i)) {
int d = N[i] - '0';
int newMask = mask & ~(1<<i);
forn(j, M)
if((j*10 + d) % M == rem)
ret += solve(newMask, j);
}
return ret;
}
void solve2() {
CL(Ans, 0);
Ans[0][0] = 1;
forab(i, 1, M-1)
Ans[0][i] = 0;
forn(mask, (1<<sz)) forn(rem, M){
forn(i, sz) if((mask & (1<<i)) == 0) {
int d = N[i] - '0';
int newMask = mask | (1<<i);
Ans[newMask][(rem*10 + d)%M] += Ans[mask][rem];
}
}
}
long long DivisiblePermutations::count (string _N, int _M)
{
sort(all(_N));
N = _N;
M = _M;
sz = N.size();
/*
CL(Ans, -1);
int64 ret = solve((1<<sz)-1, 0);
*/
solve2();
int64 ret = Ans[(1<<sz)-1][0];
DEBUG(ret);
forab(i, 1, 9) {
int cnt = 0;
tr(it, N )
cnt += *it == i + '0' ? 1 : 0;
ret /= fact(cnt);
}
return ret;
}
// BEGIN KAWIGIEDIT TESTING
// Generated by KawigiEdit 2.1.8 (beta) modified by pivanof
#include <iostream>
#include <string>
#include <vector>
using namespace std;
bool KawigiEdit_RunTest(int testNum, string p0, int p1, bool hasAnswer, long long p2) {
cout << "Test " << testNum << ": [" << "\"" << p0 << "\"" << "," << p1;
cout << "]" << endl;
DivisiblePermutations *obj;
long long answer;
obj = new DivisiblePermutations();
clock_t startTime = clock();
answer = obj->count(p0, p1);
clock_t endTime = clock();
delete obj;
bool res;
res = true;
cout << "Time: " << double(endTime - startTime) / CLOCKS_PER_SEC << " seconds" << endl;
if (hasAnswer) {
cout << "Desired answer:" << endl;
cout << "\t" << p2 << endl;
}
cout << "Your answer:" << endl;
cout << "\t" << answer << endl;
if (hasAnswer) {
res = answer == p2;
}
if (!res) {
cout << "DOESN'T MATCH!!!!" << endl;
} else if (double(endTime - startTime) / CLOCKS_PER_SEC >= 2) {
cout << "FAIL the timeout" << endl;
res = false;
} else if (hasAnswer) {
cout << "Match :-)" << endl;
} else {
cout << "OK, but is it right?" << endl;
}
cout << "" << endl;
return res;
}
int main() {
bool all_right;
all_right = true;
string p0;
int p1;
long long p2;
{
// ----- test 0 -----
p0 = "133";
p1 = 7;
p2 = 1ll;
all_right = KawigiEdit_RunTest(0, p0, p1, true, p2) && all_right;
// ------------------
}
{
// ----- test 1 -----
p0 = "2753";
p1 = 5;
p2 = 6ll;
all_right = KawigiEdit_RunTest(1, p0, p1, true, p2) && all_right;
// ------------------
}
{
// ----- test 2 -----
p0 = "1112225";
p1 = 5;
p2 = 20ll;
all_right = KawigiEdit_RunTest(2, p0, p1, true, p2) && all_right;
// ------------------
}
{
// ----- test 3 -----
p0 = "123456789";
p1 = 17;
p2 = 21271ll;
all_right = KawigiEdit_RunTest(3, p0, p1, true, p2) && all_right;
// ------------------
}
{
// ----- test 4 -----
p0 = "987654321999999";
p1 = 39;
p2 = 19960896ll;
all_right = KawigiEdit_RunTest(4, p0, p1, true, p2) && all_right;
// ------------------
}
if (all_right) {
cout << "You're a stud (at least on the example cases)!" << endl;
} else {
cout << "Some of the test cases had errors." << endl;
}
return 0;
}
// PROBLEM STATEMENT
// We define a permutation of an integer N as an integer that has the exact same digits as N, but possibly in a different order. Two permutations of N are considered different if the numbers they represent are not the same. For example, the set of all different permutations of the number N = 313 is {133, 313, 331}.
// Given a string N and an int M, determine the number of different permutations of N that are divisible by M.
//
//
// DEFINITION
// Class:DivisiblePermutations
// Method:count
// Parameters:string, int
// Returns:long long
// Method signature:long long count(string N, int M)
//
//
// CONSTRAINTS
// -N will contain between 1 and 15 non-zero digits ('1'-'9'), inclusive.
// -M will be between 1 and 50, inclusive.
//
//
// EXAMPLES
//
// 0)
// "133"
// 7
//
// Returns: 1
//
// There are three permutations of 133 (133, 313, 331), but only 133 is divisible by 7.
//
// 1)
// "2753"
// 5
//
// Returns: 6
//
// The permutations of 2753 that are divisible by 5 are 2375, 2735, 3275, 3725, 7235 and 7325.
//
// 2)
// "1112225"
// 5
//
// Returns: 20
//
// 3)
// "123456789"
// 17
//
// Returns: 21271
//
// 4)
// "987654321999999"
// 39
//
// Returns: 19960896
//
// END KAWIGIEDIT TESTING
//Powered by KawigiEdit 2.1.8 (beta) modified by pivanof!