D - 11
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
3 1 2 1 3
Sample Output 1
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.
Sample Input 2
1 1 1
Sample Output 2
1 1
There is one subsequence with length 1: 1.
There is one subsequence with length 2: 1,1.
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; }