이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
const int MAX=1000005;
int dp[MAX][3],mod;
int add(int a,int b){
return a+b-(a+b>=mod?mod:0);
}
int solve1(int n,int d){
if(d==-2||d==2)return 0;
if(!n)return 1;
int&ret=dp[n][d+1];
if(~ret)return ret;
ret=add(solve1(n-1,d-1),solve1(n-1,d+1));
return ret;
}
int solve2(int n,int d){
if(d==-1||d==3)return 0;
if(!n)return 1;
int&ret=dp[n][d];
if(~ret)return ret;
ret=add(solve2(n-1,d-1),solve2(n-1,d+1));
return ret;
}
int solve3(int n,int d){
if(d==-3||d==1)return 0;
if(!n)return 1;
int&ret=dp[n][-d];
if(~ret)return ret;
ret=add(solve3(n-1,d-1),solve3(n-1,d+1));
return ret;
}
int index(string&s){
int mn=0,mx=0,d=0,res=0,n=s.size();
for(int i=0;i<n;i++){
if(s[i]=='L'){
mn=min(mn,d-1);
d--;
}else{
int x=0;
if(mx<=1&&min(mn,d-1)>=-1)res=add(res,solve1(n-i-1,d-1)),x+=solve1(n-i-1,d-1)>=1; // -1 to 1
if(mx<=2&&min(mn,d-1)>=0)res=add(res,solve2(n-i-1,d-1)),x+=solve2(n-i-1,d-1)>=1; // 0 to 2
if(mx<=0&&min(mn,d-1)>=-2)res=add(res,solve3(n-i-1,d-1)),x+=solve3(n-i-1,d-1)>=1;
if(x>1)res=add(res,mod-1);
mx=max(mx,d+1);
d++;
}
}
return add(res,1);
}
int main(){
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
memset(dp,-1,sizeof(dp));
int n;
string s;
cin>>n>>mod>>s;
cout<<index(s);
return 0;
}
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