oj.uz

답안 #19019

# 제출 시각 아이디 문제 언어 결과 실행 시간 메모리
19019 2016-02-17T04:34:46 Z tncks0121 카드 (kriii4_Z) C++14
100 / 100
 181 ms 5708 KB 
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <memory.h>
#include <math.h>
#include <assert.h>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <algorithm>
#include <string>
#include <functional>
#include <vector>
#include <deque>
#include <utility>
#include <bitset>
#include <limits.h>
#include <time.h>
#include <functional>
#include <numeric>
#include <iostream>
#include <unordered_map>
 
using namespace std;
typedef long long ll;
typedef unsigned long long llu;
typedef double lf;
typedef unsigned int uint;
typedef long double llf;
typedef pair<int, int> pii;
typedef pair<ll, int> pli;
 
#define debug(format, ...) printf(format, __VA_ARGS__);
 
const ll MOD = (ll)1e9 + 7;
 
ll modpow (ll a, ll b) {
    a %= MOD;
    ll ret = 1;
    while(b > 0) {
        if(b & 1) ret = (ret * a) % MOD;
        a = (a * a) % MOD;
        b >>= 1;
    }
    return ret;
}
 
struct mint {
    ll val;
    mint(ll val = 0): val((val % MOD + MOD) % MOD) { }
    mint operator+(mint p) { return val + p.val; }
    mint operator-(mint p) { return val - p.val; }
    mint operator*(mint p) { return val * p.val; }
    mint operator/(mint p) { return val * modpow(p.val, MOD-2); }
};


int L;
 
const int N_ = 3050, L_ = 3050;
 
int D[N_];
int freq[N_];
 
mint invfac[L_], inv[L_], fac[L_];

#define REP(i, a, b) for (int i = (a), _end_ = (b); i < _end_; ++i)
#define mp make_pair
#define x first
#define y second
#define pb push_back
#define SZ(x) (int((x).size()))
#define ALL(x) (x).begin(), (x).end()

template<typename T> inline bool chkmin(T &a, const T &b) { return a > b ? a = b, 1 : 0; }
template<typename T> inline bool chkmax(T &a, const T &b) { return a < b ? a = b, 1 : 0; }

typedef long long LL;

const int oo = 0x3f3f3f3f;

const int Mod = 1e9 + 7;

const int maxk = 3050*2, max0 = 1<<15, max1 = 30;


inline int fpm(LL b, int e, int m)
{
	b %= m;
	LL t = 1;
	for ( ; e; e >>= 1, (b *= b) %= m)
		if (e & 1) (t *= b) %= m;
	return t;
}

int ans;
int MXN;

struct comp
{
	double x, y;

	comp(): x(0), y(0) { }
	comp(const double &_x, const double &_y): x(_x), y(_y) { }

};

inline comp operator+(const comp &a, const comp &b) { return comp(a.x + b.x, a.y + b.y); }
inline comp operator-(const comp &a, const comp &b) { return comp(a.x - b.x, a.y - b.y); }
inline comp operator*(const comp &a, const comp &b) { return comp(a.x * b.x - a.y * b.y, a.x * b.y + a.y * b.x); }
inline comp conj(const comp &a) { return comp(a.x, -a.y); }

const double PI = acos(-1);

comp w[max0 + 5];
int bitrev[max0 + 5];

int LLLL;

void fft(comp *a, const int &n)
{
	REP(i, 0, n) if (i < bitrev[i]) swap(a[i], a[bitrev[i]]);
	for (int i = 2, lyc = n >> 1; i <= n; i <<= 1, lyc >>= 1)
		for (int j = 0; j < n; j += i)
		{
			comp *l = a + j, *r = a + j + (i >> 1), *p = w;
			REP(k, 0, i >> 1)
			{
				comp tmp = *r * *p;
				*r = *l - tmp, *l = *l + tmp;
				++l, ++r, p += lyc;
			}
		}
}

inline void fft_prepare()
{
	REP(i, 0, MXN) bitrev[i] = bitrev[i >> 1] >> 1 | ((i & 1) << (LLLL - 1));
	REP(i, 0, MXN) w[i] = comp(cos(2 * PI * i / MXN), sin(2 * PI * i / MXN));
}

inline void conv(int *x, int *y, int *z)
{
	REP(i, 0, L + 1) (x[i] += Mod) %= Mod, (y[i] += Mod) %= Mod;
	static comp a[max0 + 5], b[max0 + 5];
	static comp dfta[max0 + 5], dftb[max0 + 5], dftc[max0 + 5], dftd[max0 + 5];

	REP(i, 0, L + 1) a[i] = comp(x[i] & 32767, x[i] >> 15);
	REP(i, 0, L + 1) b[i] = comp(y[i] & 32767, y[i] >> 15);
	REP(i, L + 1, MXN) a[i] = b[i] = comp(0, 0);
	fft(a, MXN), fft(b, MXN);
	REP(i, 0, MXN)
	{
		int j = (MXN - i) & (MXN - 1);
		static comp da, db, dc, dd;
		da = (a[i] + conj(a[j])) * comp(0.5, 0);
		db = (a[i] - conj(a[j])) * comp(0, -0.5);
		dc = (b[i] + conj(b[j])) * comp(0.5, 0);
		dd = (b[i] - conj(b[j])) * comp(0, -0.5);
		dfta[j] = da * dc;
		dftb[j] = da * dd;
		dftc[j] = db * dc;
		dftd[j] = db * dd;
	}
	REP(i, 0, MXN) a[i] = dfta[i] + dftb[i] * comp(0, 1);
	REP(i, 0, MXN) b[i] = dftc[i] + dftd[i] * comp(0, 1);
	fft(a, MXN), fft(b, MXN);
	REP(i, 0, L + 1)
	{
		int da = (LL)(a[i].x / MXN + 0.5) % Mod;
		int db = (LL)(a[i].y / MXN + 0.5) % Mod;
		int dc = (LL)(b[i].x / MXN + 0.5) % Mod;
		int dd = (LL)(b[i].y / MXN + 0.5) % Mod;
		z[i] = (da + ((LL)(db + dc) << 15) + ((LL)dd << 30)) % Mod;
	}
}

static int tmp[maxk + 5], a[maxk + 5], b[maxk + 5];


int main() {
	int N;
    assert(scanf("%d%d", &N, &L) == 2);
	MXN = (L + 1) << 1;
	LLLL = 0;
	while ((1 << LLLL) < MXN) ++LLLL;
	MXN = 1 << LLLL;
	fft_prepare();

    assert(1 <= N && N <= 3000);
    assert(1 <= L && L <= 3000);
    for(int i = 1; i <= N; i++) {
        assert(scanf("%d", D+i) == 1);
        assert(0 <= D[i] && D[i] <= 10);
        ++freq[D[i]];
    }

 	// precalc section
 	{
	    inv[1] = 1;
	    for(int i = 2; i < L_; i++) {
	        inv[i] = inv[MOD % i] * -(MOD / i);
	    }
	 
	    invfac[0] = fac[0] = 1;
	    for(int i = 1; i <= L; i++) {
	        fac[i] = fac[i-1] * i;
	        invfac[i] = invfac[i-1] * inv[i];
	    }
 	}
 
    vector<mint> tb(L+1, 0);
    tb[0] = 1;

    for(int d = 0; d <= 10; d++) if(freq[d] > 0) {
        vector<mint> base(L+1, 0);
        for(int j = d; j <= L; j++) base[j] = invfac[j];
 
        auto multiply = [](vector<mint> &p, vector<mint> &q) {
        	for(int i = 0; i <= L; i++) a[i] = p[i].val, b[i] = q[i].val;
        	conv(a, b, tmp);
        	vector<mint> ret(L+1);
        	for(int i = 0; i <= L; i++) ret[i] = tmp[i];
		    return ret;
        };
 
        vector<mint> coef, cur(L+1, 0);
        for(int j = d; j <= L; j++) cur[j] = invfac[j];
 
        for(int k = 0; (1<<k) <= freq[d]; k++) {
            if((freq[d] >> k) & 1) {
                if(coef.empty()) coef = cur;
                else coef = multiply(coef, cur);
            }
            cur = multiply(cur, cur);
        }
 
 		tb = multiply(coef, tb);
    }
 
    mint ans = tb[L] * fac[L] / modpow(N, L);
    printf("%lld\n", ans.val);
 
    return 0;
}

Subtask #1
13.0 / 13.0

# 결과 실행 시간 메모리 Grader output
1 Correct 52 ms 5708 KB Output is correct
2 Correct 55 ms 5708 KB Output is correct
3 Correct 47 ms 5708 KB Output is correct
4 Correct 54 ms 5708 KB Output is correct
5 Correct 50 ms 5708 KB Output is correct
6 Correct 50 ms 5708 KB Output is correct
7 Correct 50 ms 5708 KB Output is correct
8 Correct 55 ms 5708 KB Output is correct
9 Correct 49 ms 5708 KB Output is correct
10 Correct 49 ms 5708 KB Output is correct
11 Correct 49 ms 5708 KB Output is correct
12 Correct 46 ms 5708 KB Output is correct
13 Correct 58 ms 5708 KB Output is correct
14 Correct 58 ms 5708 KB Output is correct
15 Correct 50 ms 5708 KB Output is correct
16 Correct 51 ms 5708 KB Output is correct
17 Correct 58 ms 5708 KB Output is correct
18 Correct 59 ms 5708 KB Output is correct
19 Correct 49 ms 5708 KB Output is correct
20 Correct 58 ms 5708 KB Output is correct
21 Correct 39 ms 5708 KB Output is correct
22 Correct 40 ms 5708 KB Output is correct
23 Correct 32 ms 5708 KB Output is correct
24 Correct 39 ms 5708 KB Output is correct
25 Correct 42 ms 5708 KB Output is correct

Subtask #2
87.0 / 87.0

# 결과 실행 시간 메모리 Grader output
1 Correct 74 ms 5708 KB Output is correct
2 Correct 70 ms 5708 KB Output is correct
3 Correct 138 ms 5708 KB Output is correct
4 Correct 163 ms 5708 KB Output is correct
5 Correct 108 ms 5708 KB Output is correct
6 Correct 158 ms 5708 KB Output is correct
7 Correct 163 ms 5708 KB Output is correct
8 Correct 152 ms 5708 KB Output is correct
9 Correct 26 ms 5708 KB Output is correct
10 Correct 56 ms 5708 KB Output is correct
11 Correct 57 ms 5708 KB Output is correct
12 Correct 73 ms 5708 KB Output is correct
13 Correct 85 ms 5708 KB Output is correct
14 Correct 77 ms 5708 KB Output is correct
15 Correct 98 ms 5708 KB Output is correct
16 Correct 63 ms 5708 KB Output is correct
17 Correct 52 ms 5708 KB Output is correct
18 Correct 165 ms 5708 KB Output is correct
19 Correct 162 ms 5708 KB Output is correct
20 Correct 181 ms 5708 KB Output is correct