Time limit : 2sec / Memory limit : 256MB
Score : 600 points
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.
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.
Input is given from Standard Input in the following format:
n a1 a2 ... an+1
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.
3 1 2 1 3
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.
1 1 1
1 1
There is one subsequence with length 1: 1.
There is one subsequence with length 2: 1,1.
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
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.
思路:显然有一个数是重复出现的,那么需要处理下重复的。假如第i和i+3个数是一样的,此时计算子序列长度为x,在i前面和i+3后面选x-1个数就是重复的部分,减去它即可
,组合数计算用逆元。
# include
using namespace std;
typedef long long LL;
const LL mod = 1e9+7;
const LL maxn = 1e5+3;
LL inv[maxn+8]={1,1}, fi[maxn+8]={1,1}, fac[maxn+8]={1,1};
int vis[maxn+8]={0};
void init()
{
for(int i=2; i<=maxn; ++i)
{
fac[i] = fac[i-1]*i%mod;
inv[i] = (mod-mod/i)*inv[mod%i]%mod;
fi[i] = fi[i-1]*inv[i]%mod;
}
}
LL c(LL n, LL m)
{
return fac[n]*fi[n-m]%mod*fi[m]%mod;
}
int main()
{
init();
int n, t, dis;
scanf("%d",&n);
for(int i=1; i<=n+1; ++i)
{
scanf("%d",&t);
if(vis[t])
dis = vis[t]+n-i;
vis[t] = i;
}
for(int i=1; i<=n+1; ++i)
{
LL ans = c(n*1LL+1, i*1LL)%mod;
if(dis >= i-1) ans = (ans-c(dis*1LL, i*1LL-1)+mod)%mod;
printf("%lld\n",ans);
}
return 0;
}