이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <iostream>
#include <algorithm>
#include <set>
using namespace std;
const int N = 1e5+7;
typedef long long ll;
const ll MOD = 1e9+7;
int A[N],base[N];
ll dp[N][2]; // 0 - prev was not split, 1 - prev was split
ll X(int n){
	if (n<0){
		return 0;
	}
	return (n+2)/2;
}
void normalize(int *A,int &n){
	multiset<int> S;
	for(int i = 1;i<=n;i+=1){
		S.insert(A[i]);
	}
	int ptr = 0;
	while(1){
		bool flag = false;
		for(auto it = S.rbegin();it!=S.rend();it = next(it)){
			int v = *it;
			if (S.count(v-1)){
				S.erase(S.lower_bound(v));
				S.erase(S.lower_bound(v-1));
				S.insert(v+1);
				flag = true;
				break;
			}
			else if (v>=2 && S.count(v)>=2){
				S.erase(S.lower_bound(v));
				S.erase(S.lower_bound(v));
				S.insert(v+1);
				S.insert(v-2);
				flag = true;
				break;
			}
			else if (S.count(v)==2 && v==1){
				S.erase(S.lower_bound(1));
				S.erase(S.lower_bound(1));
				S.insert(0);
				S.insert(2);
				flag = true;
				break;
			}
			else if (S.count(v)==2 && v==0){
				S.erase(S.lower_bound(0));
				S.erase(S.lower_bound(0));
				S.insert(1);
				flag = true;
				break;
			}
		}
		if (!flag){
			break;
		}
	}
	
	for(int to:S){
		A[++ptr] = to;
	}
	n = ptr;
}
ll solve(int &n){
	normalize(A,n);
	dp[0][0] = 1;
	A[0] = -1;
	for(int i = 1;i<=n;i+=1){
		dp[i][0] = (A[i-1]!=A[i]?dp[i-1][0]:0)+dp[i-1][1];
		dp[i][1] = dp[i-1][0]*X(A[i]-2-A[i-1]-1)+dp[i-1][1]*X(A[i]-2-(A[i-1]-1)-1);
		dp[i][0] %= MOD;
		dp[i][1] %= MOD;
	}
	return (dp[n][0]+dp[n][1])%MOD;
}
int main(){
	int n;
	cin>>n;
	int sz = 0;
	for(int i = 1;i<=n;i+=1){
		cin>>A[++sz];
		A[sz] -= 1;
		cout<<solve(sz)<<'\n';
	/*	cout<<"GDB ";
		for(int i = 1;i<=sz;i+=1){
			cout<<A[i]<<' ';
		}
		cout<<'\n';*/
	}
}
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