abc066d <组合>

D - 11

// https://atcoder.jp/contests/abc066/tasks/arc077_b
// <组合数>
// 总组合数减去重复部分
// 对于本题求组合数方法:
// 1. 递推, 如本代码
// 2. 预处理 1e5 内的阶乘和逆元 (递推求得) (见下面第二份代码)
#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
using LL = long long;
const int N = 1e5 + 10, mod = 1e9 + 7;
int a[N];

int qpow(int a, int k)
{
    int res = 1;
    while (k)
    {
        if (k & 1) res = 1ll * res * a % mod;
        k >>= 1;
        a = 1ll * a * a % mod;
    }
    return res;
}

int C(int n, int m)
{
    if (m < 0 || m > n) return 0;
    int res = 1;
    for (int i = 1, j = n; i <= m; i ++, j --)
    {
        res = 1ll * res * j % mod;
        res = 1ll * res * qpow(i, mod-2) % mod;
    }
    return res;
}

void solv()
{
    int n;
    cin >> n;
    int l, r;
    for (int i = 1, t; i <= n+1; i ++)
    {
        cin >> t;
        if (a[t]) // 记录两个相同数字的位置
        {
            l = a[t];
            r = i;
        }
        a[t] = i;
    }

    cout << n << endl;
    int t1, t2;
    for (int i = 2; i <= n+1; i ++)
    {
        if (i==2)  
        {
            t1 = C(n+1, i);
            t2 = C(n+1-(r-l+1), i-1);
        }
        else  // 采用递推计算组合数, 直接计算超时
        {
            t1 = 1ll * t1 * (n+1-i+1) % mod * qpow(i, mod-2) % mod;
            t2 = 1ll * t2 * (n+1-(r-l+1) - (i-1) + 1) % mod * qpow(i-1, mod - 2) % mod;
        }
        int ans = (1ll*t1 - t2 + mod) % mod;
        cout << ans << endl;
    }
}

int main()
{
    ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    int T = 1;
	// cin >> T;
    while (T --)
    {
        solv();
    }
    return 0;
}

递推预处理fact inv, 计算组合数

const int mod = 1e9 + 7;
const int maxn = 1e5 + 5;

int fact[maxn];
int inv[maxn];

int qpow(int a, int k)
{
    int res = 1;
    while (k)
    {
        if (k & 1) res = 1ll * res * a % mod;
        k >>= 1;
        a = 1ll * a * a % mod;
    }
    return res;
}


void pre(int n)
{
	fact[0] = 1;
	for (int i = 1; i <= n; ++i)
		fact[i] = 1LL * fact[i - 1] * i % mod;
	inv[n] = qpow(fact[n], mod - 2);
	for (int i = n; i >= 0; --i)
		inv[i - 1] = 1LL * inv[i] * i % mod;
}

int C(int n, int k) { return n < k ? 0 : 1LL * fact[n] * inv[n - k] % mod * inv[k] % mod; }