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
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，其实就是给你1到n的数，另外再加一个1到n的数。
思路 假设没有重复的点，C（n+1,k）k取1~n+1，设i，j重复（j>i） 可以推出 有重复的C（n-(j-i),k-1）,所以数量为C（n+1,k）-C（n-（j-i）,k-1）;
组合数C（n，m）=n!/(m!*(n-m)!),而这里需要mod 所以想到乘法逆元inv[i]=(i^mod-2)%mod;1/m!=(m^mod-2)%mod;然后用快速幂求。。。
#include<stdio.h>
#include<queue>
#include<math.h>
#include<time.h>
#include<string.h>
#include<vector>
#include<algorithm>
#include<iostream>
#include<set>
#include<map>
#include<stack>
#define LL long long
#define mem(a,b) memset(a,b,sizeof(a))
#define lowbit(a) a&(-a)
#define PI acos(-1)
#define shortime(a) std::ios::sync_with_stdio(a);
using namespace std;
const LL mod=1e9+7;
//long long cmp(node a,node b){ if(a.x==b.x) return a.r>b.r;return a.x>b.x;}
int maxn (int a,int b,int c){return max(max(a,b),max(b,c));}
LL min(LL a,LL b) {return a<b?a:b;}
int gcd (int a,int b){return b==0?a:gcd(b,a%b);}
LL quick(LL x,LL n){ LL ans=1,temp=x; while(n){if(n%2==1){ ans=(ans*temp)%mod;} n/=2;temp=temp*temp%mod;}return ans;}
LL inv[100005]={1,1},f1[100005]={1,1},f2[100005]={1,1};
void init()
{
for(int i=2;i<=100005;i++)
{
f1[i]=(f1[i-1]*i)%mod;
inv[i]=quick(i,mod-2);
f2[i]=(f2[i-1]*inv[i])%mod;
//printf("%d %lld %lld %lld \n",i,f1[i],inv[i],f2[i]);
}
}
LL C(LL n,LL m)
{
return f1[n]*f2[n-m]%mod*f2[m]%mod;
}
int n,vis[100005];
int main()
{
init();
scanf("%d",&n);
mem(vis,0);
LL temp;
for(int i=1;i<=n+1;i++)
{
int x;
scanf("%d",&x);
if(vis[x])
{
temp=n-(i-vis[x]);
}
vis[x]=i;
}
LL sum;
// printf()
for(int i=1;i<=n+1;i++)
{
sum=C((LL) (n+1),(LL) (i))%mod;
//printf("%lld\n",sum);
if(temp>=i-1)
{
sum=(sum-C((LL)temp,(LL)(i-1))+mod)%mod;
}
printf("%lld\n",sum);
}
return 0;
}