Submission #496385

#	Time	Username	Problem	Language	Result	Execution time	Memory
496385		XBoRickie	PIN (CEOI10_pin)	C++14	100 / 100	140 ms	18560 KiB

pin

#include <bits/stdc++.h>
using namespace std;
// Typedef
typedef long double                ld;
typedef long long int              int64;
typedef unsigned long long int     uint64;
typedef std::pair<int, int>        PII;
typedef std::pair<int64, int64>    PLL;
typedef std::vector<int>           VI;
typedef std::vector<long long>     VLL;
// Define For-loop
#define FOR(i, j, k, in)           for (int i = (j); i < (k) ; i += (in))
#define FORW(i, j, k, in)          for (int i = (j); i <= (k); i += (in))
#define RFOR(i, j, k, in)          for (int i = (j); i >= (k); i -= (in))
// Define Data structure func
#define all(cont)                  cont.begin(), cont.end()
#define rall(cont)                 cont.rbegin(), cont.rend()
#define sz(cont)                   int((cont).size())
#define pb                         push_back
#define mp                         make_pair
#define fi                         first
#define se                         second
// Define number
#define IINF                       0x3f3f3f3f
#define LLINF                      1000111000111000111LL
#define PI                         3.1415926535897932384626433832795
// Other
#define lend                       '\n'
#define hardio(name)               freopen(name".inp","r",stdin), freopen(name".out","w",stdout);
void FastIO() { std::ios_base::sync_with_stdio(false); std::cin.tie(NULL); cin.exceptions(cin.failbit); srand(time(NULL)); }

const int N = 5e5 + 6, base = 331, LG = (1 << 4) - 1;
const int MOD = 1e9 + 7, MOD2 = 1e9 + 9;
// ======================================================================

int n, d;

map<int, int> hsh;
int cnt[56] = {};
int64 val[6] = {};
string s[N] = {};

/*
    abcd
    => 4: abcd = 1
    => 3: abc abd acd bcd = 4
    => 2: ab ac ad bc bd cb = 6
    => 1: a b c d = 4
    ======================
    We can see the answer will be overlapping:
        + With 3 characters, there'll be 4C3 way to abstract [choose 4]
        + With 2 characters, there'll be 4C2 way to abstract [choose 4], 3C2 way to abstract [choose 3]
        + With 1 characters, there'll be 4C1 way to abstract [choose 4], 3C1 way to abstract [choose 3], 2C1 way to abstract [choose 2]
        + With 0 characters, 4C0 + 3C0 + 2C0 + 1C0
*/

int main(int argc, char* argv[]) { FastIO();

    cin >> n >> d;
    FORW(i, 1, n, 1) { cin >> s[i]; s[i] = '$' + s[i]; }
    FORW(i, 1, LG, 1) cnt[i] = cnt[i >> 1] + (i & 1);

    FORW(mask, 0, LG, 1) {
        hsh.clear();
        FORW(i, 1, n, 1) {
            int sum = 0;
            FORW(k, 1, 4, 1) sum = sum * base + ((mask >> (k - 1)) & 1 ? s[i][k] : 1);
            val[cnt[mask]] += hsh[sum];
            hsh[sum]++;
        }
    }

    val[3] -= 4 * val[4];
    val[2] -= 6 * val[4] + 3 * val[3];
    val[1] -= 4 * val[4] + 3 * val[3] + 2 * val[2];
    val[0] -= val[1] + val[2] + val[3] + val[4];

    cout << val[4 - d] << '\n';

    return 0; }

• Subtask 1: 100 / 100