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.

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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=200005;
const int maxn = 1e5 + 10;
ll pos[maxn],fac[maxn],facm[maxn];
ll quick_pow(ll a,ll n,ll p)
{
    ll x = a;
    ll res = 1;
    while(n){
        if(n & 1){
            res = ((ll)res * (ll)x) % p;
        }
        n >>= 1;
        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[0] = 1;
    for(int i = 1;i < maxn;i++)
        fac[i] = (fac[i-1]*i)%MOD;
    ll n;
    scanf("%lld", &n);
    n++;
    ll m, x;
    for(int i = 1;i <= n;i++){
        scanf("%lld", &x);
        if (pos[x]){
            m = n - (i - pos[x] + 1);
            break;
        }
        pos[x] = i;
    }
    for(int i = 1;i <= n;i++)
        printf("%lld
", (C(n, i) - C(m, i-1)+MOD) % MOD);
    return 0;
}