AtCoder Regular Contest 077 D

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

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.

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.

Sample Input 1

Copy

3

1 2 1 3

Sample Output 1

Copy

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.

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.

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

Be sure to print the numbers modulo 109+7.

题目大意：从n+1个序列中选出长度为k的不同子序列的个数
解题思路：
注意到有两个相同的元素
当子序列不含有相同的元素中间的元素时，子序列会被多算一次，其他情况确定，所以最终结果为：
（C（n+1，k）-C（n-d，k-1））%MOD

 #include <iostream>

 #include <algorithm>

 #include <cstring>

 #include <cstdio>

 #include <vector>

 #include <cstdlib>

 #include <iomanip>

 #include <cmath>

 #include <ctime>

 #include <map>

 #include <set>

 #include <queue>

 #include <stack>

 using namespace std;

 #define lowbit(x) (x&(-x))

 #define max(x,y) (x>y?x:y)

 #define min(x,y) (x<y?x:y)

 #define MAX 100000000000000000

 #define MOD 1000000007

 #define pi acos(-1.0)

 #define ei exp(1

 #define PI 3.141592653589793238462

 #define INF 0x3f3f3f3f3f

 #define mem(a) (memset(a,0,sizeof(a)))

 typedef long long ll;

 ll gcd(ll a,ll b){

     return b?gcd(b,a%b):a;

 }

 const int N=;

 const int maxn = 1e5 + ;

 ll pos[maxn],fac[maxn],facm[maxn];

 ll quick_pow(ll a,ll n,ll p)

 {

     ll x = a;

     ll res = ;

     while(n){

         if(n & ){

             res = ((ll)res * (ll)x) % p;

         }

         n >>= ;

         x = ((ll)x*(ll)x) % p;

     }

     return res;

 }

 ll C(ll n,ll k){

     if(k > n) return 0ll;

     ll ans = fac[k]*fac[n-k]%MOD;

     ans = (fac[n]*quick_pow(ans,MOD-2ll,MOD))%MOD;

     return ans;

 }

 int main()

 {

     fac[] = ;

     for(int i = ;i < maxn;i++)

         fac[i] = (fac[i-]*i)%MOD;

     ll n;

     scanf("%lld", &n);

     n++;

     ll m, x;

     for(int i = ;i <= n;i++){

         scanf("%lld", &x);

         if (pos[x]){

             m = n - (i - pos[x] + );

             break;

         }

         pos[x] = i;

     }

     for(int i = ;i <= n;i++)

         printf("%lld\n", (C(n, i) - C(m, i-)+MOD) % MOD);

     return ;

 }