//解復元いらなくないですか#include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #define X first #define Y second #define pb push_back #define rep(X,Y) for (int (X) = 0;(X) < (Y);++(X)) #define rrep(X,Y) for (int (X) = (Y)-1;(X) >=0;--(X)) #define repe(X,Y) for ((X) = 0;(X) < (Y);++(X)) #define peat(X,Y) for (;(X) < (Y);++(X)) #define all(X) (X).begin(),(X).end() #define rall(X) (X).rbegin(),(X).rend() #define eb emplace_back #define UNIQUE(X) (X).erase(unique(all(X)),(X).end()) using namespace std; typedef long long ll; typedef pair pii; typedef pair pll; template using vv=vector>; template ostream& operator<<(ostream &os, const vector &t) { os<<"{"; rep(i,t.size()) {os< ostream& operator<<(ostream &os, const pair &t) { return os<<"("< int main(){ ios_base::sync_with_stdio(false); cout<>n>>x; vector a(n); rep(i,n){ cin>>a[i].X; a[i].Y=i; } vector inv(n); for(pll p:a) inv[p.Y]=p.X; sort(all(a)); ll T=1e5,t=n; map st[112]; st[0][0]=-1; rep(i,n){ if(a[i].X>=T){ t=i; break; } st[i+1]=st[i]; for(pll x:st[i]) st[i+1][x.X+a[i].X]=i; } rep(i,1<<(n-t)){ ll sum=0; rep(j,n-t) if(i>>j&1) sum+=a[t+j].X; if(st[t].count(x-sum)){ string str(n,'x'); rep(j,n-t) if(i>>j&1) str[a[t+j].Y]='o'; ll rem=x-sum; for(int d=t; d>=0; --d){ d=st[d][rem]; str[a[d].Y]='o'; rem-=a[d].X; } cout<