题意:给定一个文章,问你有多少种读法,计算方法有两种,如果在$中,如果有多个空格就算n+1,如果是一个就算2的次方,如果在{}中,
那么就是把每个空格数乘起来。
析:直接模拟,每次计算一行,注意上一行最后有空格,下面第一个也是,要全部算上。
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> #define debug puts("+++++") //#include <tr1/unordered_map> #define freopenr freopen("in.txt", "r", stdin) #define freopenw freopen("out.txt", "w", stdout) using namespace std; //using namespace std :: tr1; typedef long long LL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; const double inf = 0x3f3f3f3f3f3f; const LL LNF = 0x3f3f3f3f3f3f; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 1e3 + 5; const LL mod = 1e9 + 7; const int N = 1e6 + 5; const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1}; const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1}; const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"}; inline LL gcd(LL a, LL b){ return b == 0 ? a : gcd(b, a%b); } inline int gcd(int a, int b){ return b == 0 ? a : gcd(b, a%b); } inline int lcm(int a, int b){ return a * b / gcd(a, b); } 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 int Min(int a, int b){ return a < b ? a : b; } inline int Max(int a, int b){ return a > b ? a : b; } inline LL Min(LL a, LL b){ return a < b ? a : b; } inline LL Max(LL a, LL b){ return a > b ? a : b; } inline bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m; } char s[1000000]; const int ttt = 100000; int main(){ while(scanf("%d", &n) == 1){ getchar(); LL ans = 1; bool ok = false; bool is = false; int cnt1 = 0, cnt2 = 0; bool ok1 = false; bool ok2 = false; while(n--){ gets(s); for(int i = 0; s[i]; ++i){ if(s[i] == '$' && !ok1) ok1 = true; else if(s[i] == '{') ok2 = true; else if(s[i] == '$'){ ok1 = false; if(!cnt1) continue; if(cnt1 == 1){ if(ans * 2LL > ttt) ok = true; else ans *= 2LL; } else { if(ans * (cnt1+1) > ttt) ok = true; else ans *= (cnt1 + 1); } cnt1 = 0; } else if(s[i] == '}'){ ok2 = false; if(ans * (cnt2+1) > ttt) ok = true; else ans *= cnt2 + 1; cnt2 = 0; } else if(s[i] == '|' && ok2) ++cnt2; else if(s[i] == ' ' && ok1) ++cnt1; else{ if(ok1 && cnt1){ if(cnt1 == 1){ if(ans * 2LL > ttt) ok = true; else ans *= 2LL; } else { if(ans * (cnt1+1) > ttt) ok = true; else ans *= (cnt1 + 1); } cnt1 = 0; } } } } if(ok) puts("doge"); else printf("%I64d ", ans); } return 0; }