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  • AtCoder Regular Contest 077 D

    D - 11


    Time limit : 2sec / Memory limit : 256MB

    Score : 600 points

    Problem Statement

    You are given an integer sequence of length n+1a1,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≤ain
    • 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

    3
    1 2 1 3
    

    Sample Output 1

    3
    5
    4
    1
    

    There are three subsequences with length 11 and 2 and 3.

    There are five subsequences with length 21,1 and 1,2 and 1,3 and 2,1 and 2,3.

    There are four subsequences with length 31,1,3 and 1,2,1 and 1,2,3 and 2,1,3.

    There is one subsequence with length 41,2,1,3.


    Sample Input 2

    1
    1 1
    

    Sample Output 2

    1
    1
    

    There is one subsequence with length 11.

    There is one subsequence with length 21,1.


    Sample Input 3

    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

    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.

    很烦,最后结果忘记取模,求逆元忘记可能为负。

    假设n+1个数互不相同,正确结果=C(n+1,i) - 实际重复的情况。

    #include<cstdio>
    #include<cstdlib>
    #include<cstring>
    #include<string>
    #include<algorithm>
    #include<iostream>
    #include<queue>
    #include<map>
    #include<cmath>
    #include<set>
    #include<stack>
    #define ll long long
    #define pb push_back
    #define max(x,y) ((x)>(y)?(x):(y))
    #define min(x,y) ((x)>(y)?(y):(x))
    #define cls(name,x) memset(name,x,sizeof(name))
    using namespace std;
    const int inf=1e9+10;
    const ll llinf=1e16+10;
    const int maxn=1e5+10;
    const int maxm=1e5+10;
    const int mod=1e9+7;
    const double pi=acos(-1.0);
    int n;
    int vis[maxn];
    int num[maxn];
    ll a,b;
    ll ca[maxn];
    ll cb[maxn];
    
    ll exgcd(ll a,ll b,ll &x,ll &y)
    {
        if(b==0)
        {
            x=1;
            y=0;
            return a;
        }
        ll r=exgcd(b,a%b,x,y);
        ll t=x;
        x=y;
        y=t-a/b*y;
        return r;
    }
    
    void cfunc(ll C[],ll x)
    {
        ll xx,yy;
        C[0]=C[x]=1;
        for(ll i=1;i<x;i++)
        {
            ll t=(C[i-1]*(x-i+1))%mod;
            exgcd(i,(ll)mod,xx,yy);
            C[i]=((t*xx)%mod+mod)%mod;
        }
    }
    int main()
    {
        //freopen("in.txt","r",stdin);
        while(~scanf("%d",&n))
        {
            cls(ca,0);
            cls(cb,0);
            cls(vis,0);
            int st,ed;
            for(int i=1;i<=n+1;i++)
            {
                scanf("%d",&num[i]);
                if(vis[num[i]]) ed=i;
                else vis[num[i]]=1;
            }
            for(int i=1;i<=n+1;i++)
            {
                if(num[i]==num[ed])
                {st=i;break;}
            }
            a=n+1;
            b=n+1-(ed-st+1);
            cfunc(ca,a);
            cfunc(cb,b);
            for(int i=1;i<=n+1;i++)
                printf("%lld
    ",((ca[i]-cb[i-1])%mod+mod)%mod);
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/mgz-/p/7119223.html
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