AtCoder Regular Contest 077 D 11（思维题）

题目链接 ： https://arc077.contest.atcoder.jp/tasks/arc077_b

D - 11

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Time limit : 2sec / Memory limit : 256MB

Score : 600 points

Problem Statement

You are given an integer sequence of length n+1, a1,a2,…,an+1, which consists of the n integers 1,…,n. It is known that each of the n integers 1,…,n appears at least once in this sequence.

For each integer k=1,…,n+1, find the number of the different subsequences (not necessarily contiguous) of the given sequence with length k, modulo 109+7.

Notes

• If the contents of two subsequences are the same, they are not separately counted even if they originate from different positions in the original sequence.

• A subsequence of a sequence a with length k is a sequence obtained by selecting k of the elements of a and arranging them without changing their relative order. For example, the sequences 1,3,5 and 1,2,3 are subsequences of 1,2,3,4,5, while 3,1,2 and 1,10,100 are not.

Constraints

• 1≤n≤105
• 1≤ai≤n
• Each of the integers 1,…,n appears in the sequence.
• n and ai are integers.

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Input

Input is given from Standard Input in the following format:

n
a1 a2 ... an+1

Output

Print n+1 lines. The k-th line should contain the number of the different subsequences of the given sequence with length k, modulo 109+7.

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Sample Input 1

Copy

3
1 2 1 3

Sample Output 1

Copy

3
5
4
1

There are three subsequences with length 1: 1 and 2 and 3.

There are five subsequences with length 2: 1,1 and 1,2 and 1,3 and 2,1 and 2,3.

There are four subsequences with length 3: 1,1,3 and 1,2,1 and 1,2,3 and 2,1,3.

There is one subsequence with length 4: 1,2,1,3.

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Sample Input 2

Copy

1
1 1

Sample Output 2

Copy

1
1

There is one subsequence with length 1: 1.

There is one subsequence with length 2: 1,1.

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Sample Input 3

Copy

32
29 19 7 10 26 32 27 4 11 20 2 8 16 23 5 14 6 12 17 22 18 30 28 24 15 1 25 3 13 21 19 31 9

Sample Output 3

Copy

32
525
5453
40919
237336
1107568
4272048
13884156
38567100
92561040
193536720
354817320
573166440
818809200
37158313
166803103
166803103
37158313
818809200
573166440
354817320
193536720
92561040
38567100
13884156
4272048
1107568
237336
40920
5456
528
33
1

Be sure to print the numbers modulo 109+7.

题意： 给你n+1个数 1-n每个数起码出现一次，也就是说最多只有一个重复的数，问你组成[1---n+1]长度的序列有多少种方案。

思想：假如没有出现那个重复的点  结果就是C(n+1,i) (i -->[1,n+1])   出现这个重复点 考虑什么情况下会出现重复的情况

假设第 j 个点和第 j 个点是相同的话。 那么 i 之前的点选 i 或者 选 j 这个点 对于 再选 j 后边的点 的到的序列是一样的，所以这样的肯定是重复的。所以只要从 i 之前和 j 之后同时选择数肯定会重复，因此用C(n+1,k)-C(i-1+n+1-j ,(k-1))  (k代表选择的长度)

所以用逆元求一波组合数就OK了，一定注意好取模

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const LL mod = 1e9+7;
int maxn;
int vis[100005];
LL fac[100005]={1,1};
LL inv[100005]={1,1};
LL fi[100005]={1,1}; 
void init()
{
    for(int i=2; i<=100005; ++i)
    {
        fac[i] = fac[i-1]*i%mod;
        inv[i] = (mod-mod/i)*inv[mod%i]%mod;
        fi[i] = fi[i-1]*inv[i]%mod;
    }
}
LL c(LL n, LL m)
{
    return fac[n]*fi[n-m]%mod*fi[m]%mod;
}
int main()
{
	init();
	int n,dist;
	scanf("%d",&n);
	for(int i=1;i<=n+1;i++)
	{
		int temp;
		scanf("%d",&temp);
		if(vis[temp]) 
			dist = vis[temp] + n - i;	
		vis[temp] = i;
	}	
	long long ans;
	for(int i=1;i<=n+1;i++)
	{
		ans = c((long long)(n+1), (long long)(i))%mod;
		if(dist>=i-1)
		{
			ans= (ans - c((long long)dist, (long long)(i-1)) + mod)%mod;
		}
		printf("%lld\n",ans);	
	}
	return 0;
}