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 = 107;
typedef long long ll;
const ll MOD = 1e9+7;
int A[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+1)/2;
}
ll solve(int &n){
	multiset<int> S;
	for(int i = 1;i<=n;i+=1){
		S.insert(A[i]);
	}
	int ptr = 0;
	while(!S.empty()){
		int v = *S.begin();
		S.erase(S.begin());
		if (v==1 && S.count(v)!=0){
			S.erase(S.lower_bound(1));
			S.insert(2);
		}
		else if (S.count(v+1)!=0){
			S.erase(S.lower_bound(v+1));
			S.insert(v+2);
		}
		else{
			A[++ptr] = v;
		}
	}
	n = ptr;
	dp[0][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])+dp[i-1][1]*X(A[i]-2-(A[i-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];
		cout<<solve(sz)<<'\n';
	}
}
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