题意：给出n个数，其中有一个数会出现两次，其余数只出现一次，问不同长度且不同的子串的数量。取模1e9+7
思路：组合求出所有情况,减去重复情况,注意用逆元即可
/** @Date : 2017-07-06 09:56:44
* @FileName: atcoder077 D 组合.cpp
* @Platform: Windows
* @Author : Lweleth (SoungEarlf@gmail.com)
* @Link : https://github.com/
* @Version : $Id$
*/
#include <bits/stdc++.h>
#define LL long long
#define PII pair
#define MP(x, y) make_pair((x),(y))
#define fi first
#define se second
#define PB(x) push_back((x))
#define MMG(x) memset((x), -1,sizeof(x))
#define MMF(x) memset((x),0,sizeof(x))
#define MMI(x) memset((x), INF, sizeof(x))
using namespace std;
const int INF = 0x3f3f3f3f;
const int N = 1e5+20;
const double eps = 1e-8;
const LL mod = 1e9 + 7;
LL a[N];
LL inv[N];
LL fa[N];
LL n;
void init()
{
fa[0] = fa[1] = 1;
inv[1] = 1;
for(LL i = 2; i < N; i++)
{
fa[i] = fa[i-1] * i % mod;
inv[i] = (mod - mod / i) * inv[mod % i] % mod;
}
inv[0] = 1;
for(int i = 1; i < N; i++)
(inv[i] *= inv[i - 1]) %= mod;
}
LL C(LL n, LL k)
{
LL ans = 0;
if(k > n)
return ans;
ans = ((fa[n] * inv[k] % mod) * inv[n - k]) % mod;
return ans;
}
int main()
{
init();
while(cin >> n)
{
map<LL, int>q;
LL p = 0;
for(int i = 1; i <= n + 1; i++)
{
scanf("%lld", a + i);
if(!q[a[i]])
q[a[i]] = i;
else
p = i;
}
for(int i = 0; i <= n; i++)
{
LL ans = 0;
ans = (ans + C(n + 1, i + 1)) % mod;
ans = (ans - C(n - p + q[a[p]], i)) % mod;
while(ans < 0)
ans += mod;
printf("%lld
", ans);
}
}
return 0;
}