HDU 4891 The Great Pan (题意题+模拟)

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\n", ans);

    }

    return 0;

}