http://acm.hdu.edu.cn/showproblem.php?pid=4891

给出一个文本，问说有多少种理解方式。

1. $$中间的，(s1+1) * (s2+1) * ...*(sn+1), si表示连续的空格数。

2.{}中间，即 | 的个数+1.

就是模拟

#include <iostream>

#include <cstring>

#include <cstdio>

#include <algorithm>

#include <vector>

#include <string>

#include <cmath>

using namespace std;

#define RD(x) scanf("%d",&x)

#define RD2(x,y) scanf("%d%d",&x,&y)

#define clr0(x) memset(x,0,sizeof(x))

typedef long long LL;

#define N 100005

int n;

string x , y;

LL check(int l , int r){

    LL ans = 1,tmp = 1;

    for (int i=l;i<=r;++i){

        if (x[i] == ' ')

            ++tmp;

        else{

           ans *= tmp;

           tmp = 1;

           if (ans > 100000)

            return -1;

        }

    }

    return ans;

}

LL checkk(int l , int r){

    LL ans = 1;

    for (int i=l;i<=r;++i)

        if (x[i] == '|') ++ans;

    return ans;

}

void work(){

     getchar();

     x = "";

     while (n--){

           getline(cin,y);

           x += y;

     }

     int m = x.size();

     LL ans = 1 , now;

     for(int i = 0;i < m;){

           while (x[i] != '$' && x[i] != '{' && i < m) ++i;

           if (i == m) break;

           int j = i+1;

           if (x[i] == '$'){

                while (x[j] != '$') ++j;

                now = check(i,j);

           } else {

              while (x[j] != '}') ++j;

              now = checkk(i,j);

           }

           if (now == -1){

               puts("doge");

               return;

           }

           ans *= now;

           if (ans > 100000){

               puts("doge");

               return;

           }

           i = j+1;

     }

     printf("%I64d\n",ans);

}

int main(){

    while (~RD(n))

        work();

    return 0;

}