# | 제출 시각 | 아이디 | 문제 | 언어 | 결과 | 실행 시간 | 메모리 |
---|---|---|---|---|---|---|---|
504078 | Carmel_Ab1 | Linear Garden (IOI08_linear_garden) | C++17 | 1578 ms | 6072 KiB |
이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
/*
#pragma GCC target ("avx2")
#pragma GCC optimization ("O3")
#pragma GCC optimization ("unroll-loops")
*/
#include<bits/stdc++.h>
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
//using namespace __gnu_pbds;
using namespace std;
typedef long double ld;
typedef long long ll;
typedef unsigned long long ull;
typedef vector<int>vi;
typedef vector<vector<int>>vvi;
typedef vector<ll>vl;
typedef vector<vl> vvl;
typedef pair<int,int>pi;
typedef pair<ll,ll> pl;
typedef vector<pl> vpl;
typedef vector<ld> vld;
typedef pair<ld,ld> pld;
typedef vector<pi> vpi;
//typedef tree<ll, null_type, less_equal<ll>,rb_tree_tag,tree_order_statistics_node_update> ordered_set;
template<typename T> ostream& operator<<(ostream& os, vector<T>& a){os<<"[";for(int i=0; i<ll(a.size()); i++){os << a[i] << ((i!=ll(a.size()-1)?" ":""));}os << "]\n"; return os;}
#define all(x) x.begin(),x.end()
#define YES out("YES")
#define NO out("NO")
#define out(x){cout << x << "\n"; return;}
#define outfl(x){cout << x << endl;return;}
#define GLHF ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL)
#define print(x){for(auto ait:x) cout << ait << " "; cout << "\n";}
#define pb push_back
#define umap unordered_map
template<typename T1, typename T2> istream& operator>>(istream& is, pair<T1, T2>& p){is >> p.first >> p.second;return is;}
template<typename T1, typename T2> ostream& operator<<(ostream& os, pair<T1, T2>& p){os <<"" << p.first << " " << p.second << ""; return os;}
void usaco(string taskname){
string fin = taskname + ".in";
string fout = taskname + ".out";
const char* FIN = fin.c_str();
const char* FOUT = fout.c_str();
freopen(FIN, "r", stdin);
freopen(FOUT, "w", stdout);
}
template<typename T>
void read(vector<T>& v){
int n=v.size();
for(int i=0; i<n; i++)
cin >> v[i];
}
template<typename T>
vector<T>UNQ(vector<T>a){
vector<T>ans;
for(T t:a)
if(ans.empty() || t!=ans.back())
ans.push_back(t);
return ans;
}
void solve();
int main(){
GLHF;
int t=1;
//cin >> t;
while(t--)
solve();
}
bool ok(string s){
int n=s.size();
vi pref(n);
for(int i=0; i<n; i++)
pref[i]=(i?pref[i-1]:0)+(s[i]=='L'?1:-1);
auto sum=[&](int l,int r){
return pref[r]-(l?pref[l-1]:0);
};
for(int i=0; i<n; i++)
for(int j=i+2; j<n; j++){
if(abs(sum(i,j))>2)return 0;
}
return 1;
}
ll calc(string t){
int n=t.size(),ans=0;
vector<string>v;
for(int i=0; i<(1<<n); i++){
string s;
for(int j=0; j<n; j++)
if(i&(1<<j))
s+='L';
else
s+='P';
if(s<=t && ok(s))
v.pb(s),ans++;
}
sort(all(v));
//print(v)
cout.flush();
return ans;
}
void solve() {
ll n,mod;
cin >> n >> mod;
string s;
cin >> s;
out(calc(s)%mod)
ll ans=0;//how many are lexic. smaller than s
vector<vvl>dp(n+1,vvl(2,vl(2)));
dp[1][0][1]=dp[1][1][0]=2;
dp[1][0][0]=dp[1][1][1]=1;
for(int i=2; i<=n; i++){
for(int c1=0; c1<2; c1++)
for(int c2=0; c2<2; c2++){
if(c1!=c2)
dp[i][c1][c2]=(dp[i-1][c2][c2]+dp[i-1][c2][c2^1])%mod;
else
dp[i][c1][c2]=dp[i-1][c2][c2^1]%mod;
}
}
// L < P !!!
for(int i=1; i<n; i++){//i is tie-breaker
if(s[i]=='L')continue;
ans=(ans+dp[n-i-1][1][(s[i-1]=='L')])%mod;
}
cout << calc(s) << endl;
if(s[0]=='P') {
ans=(ans+dp[n-2][1][0])%mod;
ans=(ans+dp[n-2][1][1])%mod;
}
ans=(ans+1)%mod;
out(ans)
}
/*
12
100000000
LPLLPLPPLPLL
*/
컴파일 시 표준 에러 (stderr) 메시지
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