题意:给定一个序列,*表示雷,1表示它旁边有一个雷,2表示它旁边有两个雷,0表示旁边没有雷,?表示未知,求有多少情况。
析:dp[i][j] 表示第 i 个放 j 状态,有多少种情况,然后很简单的DP就可以搞定。
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #include <set> #include <queue> #include <algorithm> #include <vector> #include <map> #include <cctype> #include <cmath> #include <stack> #include <sstream> #define debug() puts("++++"); #define gcd(a, b) __gcd(a, b) #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define freopenr freopen("in.txt", "r", stdin) #define freopenw freopen("out.txt", "w", stdout) using namespace std; typedef long long LL; typedef unsigned long long ULL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; const LL LNF = 1e16; const double inf = 0x3f3f3f3f3f3f; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 1000000 + 10; const int mod = 1000000007; const int dr[] = {-1, 0, 1, 0}; const int dc[] = {0, 1, 0, -1}; const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"}; int n, m; const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; inline bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m; } LL dp[2][5]; char s[maxn]; /* 0 - 0 1 - *1 2 - 1* 3 - *2* 4 - * */ int main(){ cin >> s+1; n = strlen(s+1); int cnt = 0; if(s[1] == '0') dp[cnt][0] = 1; else if(s[1] == '1') dp[cnt][2] = 1; else if(s[1] == '*') dp[cnt][4] = 1; else if(s[1] == '?') dp[cnt][0] = dp[cnt][2] = dp[cnt][4] = 1; cnt ^= 1; for(int i = 2; i <= n; ++i, cnt ^= 1){ memset(dp[cnt], 0, sizeof dp[cnt]); if(s[i] == '0') dp[cnt][0] = (dp[cnt^1][0] + dp[cnt^1][1]) % mod; else if(s[i] == '1'){ dp[cnt][1] = dp[cnt^1][4]; dp[cnt][2] = (dp[cnt^1][0] + dp[cnt^1][1]) % mod; } else if(s[i] == '2') dp[cnt][3] = dp[cnt^1][4]; else if(s[i] == '*') dp[cnt][4] = (dp[cnt^1][2] + dp[cnt^1][3] + dp[cnt^1][4]) % mod; else { dp[cnt][0] = (dp[cnt^1][0] + dp[cnt^1][1]) % mod; dp[cnt][1] = dp[cnt^1][4]; dp[cnt][2] = (dp[cnt^1][0] + dp[cnt^1][1]) % mod; dp[cnt][3] = dp[cnt^1][4]; dp[cnt][4] = (dp[cnt^1][2] + dp[cnt^1][3] + dp[cnt^1][4]) % mod; } } LL ans = dp[cnt^1][0] + dp[cnt^1][1] + dp[cnt^1][4]; cout << ans % mod << endl; return 0; }