AGC022E - Median Replace

AGC022E - Median Replace

题面描述

Solution

直接构造自动机，模拟每次在末尾加一个字符的当前状态的方案数$f[state]$，状态很少，直接打表转移即可。

时间复杂度$O(n)$。

#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <ctime>
#include <cassert>
#include <string.h>
//#include <unordered_set>
//#include <unordered_map>
//#include <bits/stdc++.h>

#define MP(A,B) make_pair(A,B)
#define PB(A) push_back(A)
#define SIZE(A) ((int)A.size())
#define LEN(A) ((int)A.length())
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define fi first
#define se second

using namespace std;

template<typename T>inline bool upmin(T &x,T y) { return y<x?x=y,1:0; }
template<typename T>inline bool upmax(T &x,T y) { return x<y?x=y,1:0; }

typedef long long ll;
typedef unsigned long long ull;
typedef long double lod;
typedef pair<int,int> PR;
typedef vector<int> VI;

const lod eps=1e-11;
const lod pi=acos(-1);
const int oo=1<<30;
const ll loo=1ll<<62;
const int mods=1e9+7;
const int MAXN=600005;
const int INF=0x3f3f3f3f;//1061109567
/*--------------------------------------------------------------------*/
inline int read()
{
	int f=1,x=0; char c=getchar();
	while (c<'0'||c>'9') { if (c=='-') f=-1; c=getchar(); }
	while (c>='0'&&c<='9') { x=(x<<3)+(x<<1)+(c^48); c=getchar(); }
	return x*f;
}
char st[MAXN];
int f[MAXN][8];
int to[8][2]={{2,1},{4,3},{6,5},{3,3},{7,1},{2,1},{2,2},{4,4}};
int upd(int x,int y){ return x+y>=mods?x+y-mods:x+y; }
int main()
{
	scanf("%s",st);
	int len=strlen(st);
	f[0][0]=1;
	for (int i=0;i<len;i++)
		for (int j=0;j<=7;j++) 
		{
			if (st[i]!='0') f[i+1][to[j][1]]=upd(f[i+1][to[j][1]],f[i][j]);
			if (st[i]!='1') f[i+1][to[j][0]]=upd(f[i+1][to[j][0]],f[i][j]);
		}
	printf("%d\n",upd(f[len][1],f[len][3]));
	return 0;
}