제출 #227137

#제출 시각아이디문제언어결과실행 시간메모리
227137anubhavdharLinear Garden (IOI08_linear_garden)C++14
95 / 100
87 ms56484 KiB
#include<bits/stdc++.h> #define ll long long int #define pb push_back #define mp make_pair #define FOR(i,n) for(i=0;i<(n);++i) #define FORe(i,n) for(i=1;i<=(n);++i) #define FORr(i,a,b) for(i=(a);i<(b);++i) #define FORrev(i,n) for(i=(n);i>=0;--i) #define ii pair<ll,ll> #define vi vector<ll> #define vii vector<ii> #define ff first #define ss second #define cd complex<double> #define vcd vector<cd> #define ldd long double #define all(x) (x).begin(),(x).end() using namespace std; const short int __PRECISION = 10; //const ll MOD = 1e9+7; const ll INF = 1e17 + 1101; const ll LOGN = 17; const ll MAXN = 1e6+5; const ll ROOTN = 320; const ldd PI = acos(-1); const ldd EPS = 1e-7; ll N,MOD,dp[MAXN][3]; vector<bool> V; ll DP(ll ind,short int sum,short int min_sum,short int max_sum) { // cout<<"In DP("<<ind<<','<<sum<<','<<min_sum<<','<<max_sum<<")\n"; assert(abs(sum) <= 2); if(abs(sum) == 2 or abs(min_sum) == 2 or abs(max_sum) == 2) return ((ind == N-1) ? 0 : DP(ind + 1,sum + (V[ind]?1:-1),min(0,min_sum + (V[ind]?1:-1)),max(0,max_sum + (V[ind]?1:-1)))); if(ind == N-1) { // cout<<"DP("<<ind<<") = "<<((V[ind])?1:0)<<endl; return ((V[ind])?1:0); } sum += V[ind]?1:-1; min_sum += V[ind]?1:-1; max_sum += V[ind]?1:-1; ll val = DP(ind + 1,sum,min((short int)0,min_sum),max((short int)0,max_sum)); short int p = max(max(abs(min_sum-2),abs(max_sum-2)),abs(sum-2)); // cout<<"DP("<<ind<<") = "<<((V[ind])?1:0)*(dp[N-1-ind][p]) + val<<" where p = "<<p<<endl; return (((V[ind])?(dp[N-1-ind][p]):0) + val)%MOD; } int main() { ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL); cin>>N>>MOD; char c; int i; dp[1][0] = 2;dp[1][1] = 2;dp[1][2] = 1; dp[2][0] = 4;dp[2][1] = 3;dp[2][2] = 2; dp[3][0] = 6;dp[3][1] = 5;dp[3][2] = 2; ll pow = 2,pow1 = 2; FOR(i,N+1) { if(i > 2 and i%2 == 1) pow = (pow*2)%MOD; if(i > 3 and i%2 == 1) pow1 = (pow1*2)%MOD; if(i > 2) dp[i+1][0] = (dp[i][0] + pow)%MOD; if(i > 3) { dp[i][1] = (dp[i-1][1] + pow1)%MOD; dp[i][2] = (dp[i-1][2]*(((i%2) == 0)?2:1))%MOD; } } // FORe(i,N) // cout<<dp[i][0]<<' '<<dp[i][1]<<' '<<dp[i][2]<<endl; FOR(i,N) { cin>>c; V.pb((c == 'P')); } cout<<DP(0,0,0,0)+1; return 0; }
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