题意
Sol
这题打cf的时候真的是脑残，自己造了个abcdad的数据开心的玩了半天一脸懵逼。。。最后还好ycr大佬给了个思路不然就凉透了。。。
首先不难看出我们最后一定可以把字符串弄成(aaaabbb)的形式，若当前位和下一位不一样就直接转就行了
注意特判一下最后一个位置
/*
*/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<vector>
#include<set>
#include<queue>
#include<cmath>
#include<tr1/unordered_map>
//#include<ext/pb_ds/assoc_container.hpp>
//#include<ext/pb_ds/hash_policy.hpp>
#define Pair pair<int, int>
#define MP(x, y) make_pair(x, y)
#define fi first
#define se second
//#define int long long
#define LL long long
#define ull unsigned long long
#define rg register
#define pt(x) printf("%d ", x);
//#define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1<<22, stdin), p1 == p2) ? EOF : *p1++)
//char buf[(1 << 22)], *p1 = buf, *p2 = buf;
//char obuf[1<<24], *O = obuf;
//void print(int x) {if(x > 9) print(x / 10); *O++ = x % 10 + '0';}
//#define OS *O++ = ' ';
using namespace std;
//using namespace __gnu_pbds;
const int MAXN = 1e5 + 10, INF = 1e9 + 10, mod = 998244353;
const double eps = 1e-9;
inline int read() {
char c = getchar();
int x = 0, f = 1;
while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();}
while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
return x * f;
}
int N, flag[MAXN];
string s, r;
string RE(string s, int pre) {
for(int i = 0; i < pre; i++) swap(s[i], s[pre - i]);
return s;
}
main() {
cin >> s;
int N = s.length();
for(int i = 0; i < N; i++) {
if(s[i] != s[i + 1])
s = RE(s, i), flag[i] = 1;
// cout << s << endl;
}
string a = s, b = RE(s, N - 1);
//cout << a << " " << b << endl;
if(flag[N - 1] && (b < a)) flag[N - 1] = 0;
for(int i = 0; i < N; i++) printf("%d ", flag[i]);
return 0;
}