Fucks
A first end of each segment by separately and then combined force combined with the proviso that each segment lexicographically string does not drop, so that the minimum guaranteed to be expressed.
Because if there is a lexicographical than in any period before merger lexicographical little behind, then merge it into this string front obviously be smaller, and therefore does not meet the minimum representation, this does not need to merge.
#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
#define full(a, b) memset(a, b, sizeof a)
#define __fastIn ios::sync_with_stdio(false), cin.tie(0)
#define range(x) (x).begin(), (x).end()
#define pb push_back
using namespace std;
typedef long long LL;
inline int lowbit(int x){ return x & (-x); }
inline int read(){
int ret = 0, w = 0; char ch = 0;
while(!isdigit(ch)){
w |= ch == '-', ch = getchar();
}
while(isdigit(ch)){
ret = (ret << 3) + (ret << 1) + (ch ^ 48);
ch = getchar();
}
return w ? -ret : ret;
}
template <typename A>
inline A lcm(A a, A b){ return a / __gcd(a, b) * b; }
template <typename A, typename B, typename C>
inline A fpow(A x, B p, C lyd){
A ans = 1;
for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd;
return ans;
}
string s;
int _;
int main(){
__fastIn;
for(cin >> _; _; _ --){
cin >> s;
vector<string> v, ans;
string t = "";
t.pb(s.front());
for(int i = 1; i < s.size(); i ++){
if((s[i] == '0' && s[i - 1] == '1')){
v.pb(t), t.clear();
}
t.pb(s[i]);
}
if(!t.empty()) v.pb(t);
bool ok = false;
while(!ok){
bool y = false;
for(int i = 0; i < v.size(); i ++){
string str = v[i];
int j = i + 1;
while(j < v.size() && v[j - 1] <= v[j]) str += v[j], j ++, y = true;
ans.pb(str), i = j - 1;
}
if(!y) ok = true;
v = ans, ans.clear();
}
for(int i = 0; i < v.size(); i ++){
cout << v[i];
if(i != v.size() - 1) cout << " ";
}
cout << endl;
}
return 0;
}