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Problem link: 719 - Glass Beads
Solution:
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#include <bits/stdc++.h> | |
using namespace std; | |
#define INF 1<<30 | |
#define endl '\n' | |
#define maxn 100005 | |
#define tc printf("Case %d: ", cs) | |
#define tcn printf("Case %d:\n", cs); | |
#define FASTIO ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0); | |
typedef long long ll; | |
const double PI = acos(-1.0); | |
#define dbg1(x) cerr << #x << " = " << x << endl; | |
#define dbg2(x, y) cerr << #x << " = " << x << ", " << #y << " = " << y << endl; | |
#define dbg3(x, y, z) cerr << #x << " = " << x << ", " << #y << " = " << y << ", " << #z << " = " << z << endl; | |
#define dbg4(w,x, y, z) cerr << #w << " = " << w << ", " <<#x << " = " << x << ", " << #y << " = " << y << ", " << #z << " = " << z << endl; | |
template < typename F, typename S > | |
ostream& operator << ( ostream& os, const pair< F, S > & p ) { | |
return os << "(" << p.first << ", " << p.second << ")"; | |
} | |
template < typename T > | |
ostream &operator << ( ostream & os, const vector< T > &v ) { | |
os << "{"; | |
for (auto it = v.begin(); it != v.end(); ++it) { | |
if ( it != v.begin() ) os << ", "; | |
os << *it; | |
} | |
return os << "}"; | |
} | |
template < typename T > | |
ostream &operator << ( ostream & os, const set< T > &v ) { | |
os << "["; | |
for (auto it = v.begin(); it != v.end(); ++it) { | |
if ( it != v.begin()) os << ", "; | |
os << *it; | |
} | |
return os << "]"; | |
} | |
template < typename F, typename S > | |
ostream &operator << ( ostream & os, const map< F, S > &v ) { | |
os << "["; | |
for (auto it = v.begin(); it != v.end(); ++it) { | |
if ( it != v.begin() ) os << ", "; | |
os << it -> first << " = " << it -> second ; | |
} | |
return os << "]"; | |
} | |
#define dbg(args...) do {cerr << #args << " : "; faltu(args); } while(0) | |
clock_t tStart = clock(); | |
#define timeStamp dbg("Execution Time: ", (double)(clock() - tStart)/CLOCKS_PER_SEC) | |
void faltu () { cerr << endl; } | |
template <typename T> | |
void faltu( T a[], int n ) { | |
for (int i = 0; i < n; ++i) cerr << a[i] << ' '; | |
cerr << endl; | |
} | |
template <typename T, typename ... hello> | |
void faltu( T arg, const hello &... rest) { cerr << arg << ' '; faltu(rest...); } | |
// Program showing a policy-based data structure. | |
#include <ext/pb_ds/assoc_container.hpp> // Common file | |
#include <ext/pb_ds/tree_policy.hpp> | |
#include <functional> // for less | |
#include <iostream> | |
using namespace __gnu_pbds; | |
using namespace std; | |
// GNU link : https://goo.gl/WVDL6g | |
typedef tree<int, null_type, less_equal<int>, rb_tree_tag, | |
tree_order_statistics_node_update> | |
new_data_set; | |
/**___________________________________________________**/ | |
//Here we present the implementation of the Duval algorithm, | |
//which will return the desired Lyndon factorization of a given string s. | |
vector<string> duval(string const& s) | |
{ | |
int n = s.size(); | |
int i = 0; | |
vector<string> factorization; | |
while (i < n) { | |
int j = i + 1, k = i; | |
while (j < n && s[k] <= s[j]) { | |
if (s[k] < s[j]) k = i; | |
else k++; | |
j++; | |
} | |
while (i <= k) { | |
factorization.push_back(s.substr(i, j - k)); | |
i += j - k; | |
} | |
} | |
return factorization; | |
} | |
int min_cyclic_string(string s) | |
{ | |
s += s; | |
int n = s.size(); | |
int i = 0, ans = 0; | |
while (i < n / 2) { | |
ans = i; | |
int j = i + 1, k = i; | |
while (j < n && s[k] <= s[j]) { | |
if (s[k] < s[j]) k = i; | |
else k++; | |
j++; | |
} | |
while (i <= k) { | |
i += j - k; | |
} | |
} | |
return ans + 1; | |
//dbg(ans); | |
//return s.substr(ans, n / 2); | |
} | |
int main() | |
{ | |
FASTIO | |
///* | |
#ifndef ONLINE_JUDGE | |
freopen("in.txt", "r", stdin); | |
freopen("out.txt", "w", stdout); | |
freopen("error.txt", "w", stderr); | |
#endif | |
//*/ | |
int t; | |
cin >> t; | |
while (t--) { | |
string s; | |
cin >> s; | |
// auto ans = duval(s); | |
//dbg(ans); | |
auto res = min_cyclic_string(s); | |
// dbg(res); | |
cout << res << endl; | |
} | |
} |
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