11(ABC066&ARC077)

问题 C: 11

时间限制: 1 Sec  内存限制: 128 MB

题目描述

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.

输入

Input is given from Standard Input in the following format:
n
a1 a2 ... an+1

输出

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.

样例输入

3
1 2 1 3

样例输出

3
5
4
1

提示

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.

题意：

给你一个数 n, 序列长度为 n + 1, 保证序列中至少有 1 - n 中所有数，即某个数出现过两次，从该序列中按顺序拿 1 - > n+1个的方案

题解：

从n + 1中拿i 个 是 C(n+1,i),但对于出现两次的数，会出现重复，若出现两次的数的位置是 l  和 r，在拿了一个出现两次的那个数，对于从 l - r 区间外选取，会出现重复，即总的减去重复的就是答案。 C(n + 1,i) - C(n + 1 - (r - l + 1),i - 1)

 

求一下组合数，加逆元  该题就解决了

code

#include <bits/stdc++.h>


using namespace std;
typedef long long ll;
const int mod = 1e9 + 7;
const int N = 1e5+10;
int vis[N];
ll inv(ll b){   return b==1||b==0?1:(mod-mod/b)*inv(mod%b)%mod; }
ll f[N];
ll C(ll n,ll m)
{
    if(n<m) return 0;
    return f[n]*inv(f[m])%mod*inv(f[n-m])%mod;
}
void init()
{
    f[0]=1;
    for(int i=1;i<N;i++)
        f[i]=f[i-1]*i%mod;
}
int main()
{
    int n,first,last,dig;
    init();
    memset(vis,0,sizeof(vis));
    scanf("%d",&n);
    for(int i = 1;i <= n+1;i++)
    {
        scanf("%d",&dig);
        if(vis[dig])
            first = vis[dig],last = i;
        else
            vis[dig] = i;
    }
    for(int i = 1;i <= n+1;i++)
    {
        //cout<<C(n+1,i)<<" "<<C(n+1 - (last - first),i - 1)<<endl;
        printf("%lld\n", (C(n+1,i) - C(n + 1 - (last - first + 1),i-1) + mod)%mod);
    }
    return 0;
}