# |
Submission time |
Handle |
Problem |
Language |
Result |
Execution time |
Memory |
75152 |
2018-09-08T13:49:41 Z |
born2rule |
Match (CEOI16_match) |
C++14 |
|
450 ms |
525312 KB |
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
#define ll long long
#define db long double
#define ii pair<int,int>
#define vi vector<int>
#define fi first
#define se second
#define sz(a) (int)(a).size()
#define all(a) (a).begin(),(a).end()
#define pb push_back
#define mp make_pair
#define FN(i, n) for (int i = 0; i < (int)(n); ++i)
#define FEN(i,n) for (int i = 1;i <= (int)(n); ++i)
#define rep(i,a,b) for(int i=a;i<b;i++)
#define repv(i,a,b) for(int i=b-1;i>=a;i--)
#define SET(A, val) memset(A, val, sizeof(A))
typedef tree<int ,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update>ordered_set ;
// order_of_key (val): returns the no. of values less than val
// find_by_order (k): returns the kth largest element.(0-based)
#define TRACE
#ifdef TRACE
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
cerr << name << " : " << arg1 << std::endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
const char* comma = strchr(names + 1, ','); cerr.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
#else
#define trace(...)
#endif
const int N=100005,L=26;
int pre[L][N];
string s;
string solve(int l,int r)
{
if(l>r) return "";
if(s[l]==s[r])
{
string ans="(";
ans+=solve(l+1,r-1);
ans+=")";
return ans;
}
int ind=pre[s[l]-'a'][r];
string ans="(";
ans+=solve(l+1,ind-1);
ans+=")";
ans+=solve(ind+1,r);
return ans;
}
int main()
{
std::ios::sync_with_stdio(false);
cin.tie(NULL) ; cout.tie(NULL) ;
cin>>s;
int n=sz(s);
s="#"+s;
stack<int> st;
rep(i,1,n+1)
{
if(st.empty() || s[st.top()]!=s[i]) st.push(i);
else st.pop();
}
if(!st.empty())
{
cout<<-1<<endl;
return 0;
}
rep(i,3,n+1)
{
rep(j,0,26)
{
if(s[i]==s[i-1] && (int)(s[i-2]-'a')==j)
pre[j][i]=i-2;
else
{
int tmp=pre[(int)(s[i]-'a')][i-1];
if(!tmp)
{
if(s[i-1]==s[i]) tmp=i-2;
else break;
}
pre[j][i]=pre[j][tmp];
}
}
}
cout<<solve(1,n)<<endl;
return 0 ;
}
# |
Verdict |
Execution time |
Memory |
Grader output |
1 |
Correct |
2 ms |
376 KB |
Output is correct |
2 |
Correct |
2 ms |
500 KB |
Output is correct |
3 |
Correct |
2 ms |
500 KB |
Output is correct |
# |
Verdict |
Execution time |
Memory |
Grader output |
1 |
Correct |
2 ms |
376 KB |
Output is correct |
2 |
Correct |
2 ms |
500 KB |
Output is correct |
3 |
Correct |
2 ms |
500 KB |
Output is correct |
4 |
Runtime error |
450 ms |
525312 KB |
Execution killed with signal 9 (could be triggered by violating memory limits) |
5 |
Halted |
0 ms |
0 KB |
- |
# |
Verdict |
Execution time |
Memory |
Grader output |
1 |
Correct |
2 ms |
376 KB |
Output is correct |
2 |
Correct |
2 ms |
500 KB |
Output is correct |
3 |
Correct |
2 ms |
500 KB |
Output is correct |
4 |
Runtime error |
450 ms |
525312 KB |
Execution killed with signal 9 (could be triggered by violating memory limits) |
5 |
Halted |
0 ms |
0 KB |
- |