Submission #19950

#	Time	Username	Problem	Language	Result	Execution time	Memory
19950		tonyjjw	순열 (kriii4_T)	C++14	100 / 100	479 ms	40784 KiB

T

#include <algorithm>
#include <assert.h>
#include <complex>
#include <ctype.h>
#include <functional>
#include <iostream>
#include <limits.h>
#include <locale.h>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <time.h>
#include <vector>
#include <unordered_set>
#include <unordered_map>

#pragma warning(disable:4996)
using namespace std;

#define mp make_pair
typedef long long ll;
typedef unsigned long long ull;
typedef double db;
typedef long double ldb;
typedef pair <int, int> pii;
typedef pair <ll, ll> pll;
typedef pair <ll, int> pli;
typedef pair <ldb, ldb> pdd;

int IT_MAX = 131072;
const ll MOD = 1000000007;
const int INF = 1034567891;
const ll LL_INF = 2234567890123456789ll;
const db PI = 3.141592653589793238;
const ldb ERR = 1E-10;

ll mul_inv(ll a, ll b)
{
	ll b0 = b, t, q;
	ll x0 = 0, x1 = 1;
	if (b == 1) return 1;
	while (a > 1) {
		q = a / b;
		t = b, b = a % b, a = t;
		t = x0, x0 = x1 - q * x0, x1 = t;
	}
	if (x1 < 0) x1 += b0;
	return x1;
}

ll F[1000050];
ll Finv[1000050];
ll dp1[1000050];
ll dp2[1000050];
ll dp3[1000050];

inline ll Comb(int x, int y) {
	ll rv = F[x];
	rv = (rv * Finv[y]) % MOD;
	rv = (rv * Finv[x - y]) % MOD;
	return rv;
}

int main() {
	int N, L, K, i;
	F[0] = 1;
	for (i = 1; i <= 1000010; i++) F[i] = (F[i - 1] * i) % MOD;
	for (i = 0; i <= 1000010; i++) Finv[i] = mul_inv(F[i], MOD);

	scanf("%d %d", &N, &K);
	K++;
	if (K == N + 1) return !printf("0\n");


	ll ans = Comb(N - K + 2, 2) * N % MOD;
	ans -= 2 * Comb(N - K + 2, 3);
	ans = (ans + 2 * MOD) % MOD;
	ans = (ans*N) % MOD;
	ans = (ans*F[N - 1]) % MOD;
	for (L = K; L < N; L++) {
		dp1[L] = Comb(N + 1, L + 1) * L % MOD;
		
		dp2[L] = Comb(N + 2, L + 2);
		dp2[L] = dp2[L] * L % MOD * (L + 1) % MOD;
		dp2[L] = (dp2[L] - dp1[L] + MOD) % MOD;

		dp3[L] = Comb(N + 3, L + 3) * L % MOD;
		dp3[L] = dp3[L] * (L + 1) % MOD * (L + 2) % MOD;
		dp3[L] = (dp3[L] - 3 * dp2[L] + 3 * MOD) % MOD;
		dp3[L] = (dp3[L] - 2 * dp1[L] + 2 * MOD) % MOD;

		ll t = Comb(L - K + 2, 2)*L%MOD;
		t -= 2 * Comb(L - K + 2, 3);
		t = (t + 2 * MOD) % MOD;
		t = (t * F[N - L - 1]) % MOD;
		t = (t * F[L-1]) % MOD;
		
		ll t2 = 0;
		t2 += (2 * N * dp1[L]) % MOD;
		t2 -= 2 * dp2[L];
		t2 += (ll)N*(N - 1) % MOD*dp1[L] % MOD;
		t2 -= (2 * N - 1) * dp2[L] % MOD;
		t2 += dp3[L];
		t2 = (t2 + 100 * MOD) % MOD;

		ans += (t*t2) % MOD;
	}

	ans = ans % MOD;

	ll sum = 0;
	for (i = 1; i + K - 1 <= N; i++) sum += (N - i - K + 2);
	sum %= MOD;
	sum *= N + 1;
	sum %= MOD;
	sum = (sum * F[N]) % MOD;

	ans = sum - ans;
	ans %= MOD;
	ans += MOD;
	ans %= MOD;

	printf("%lld\n", ans);
	return 0;
}

• Subtask 1: 8 / 8
• Subtask 2: 92 / 92