M:翻译布尔表达式

翻译布尔表达式

Time Limit: 1000 ms Memory Limit: 65536 KiB

Submit Statistic

Problem Description

大家都学过了布尔表达式的翻译，其中有一个拉链－回填技术，这次我们就练习这个技术。

Input

 多组输入，首先输入一个整数T，代表数据组数。

接下来每组输入为一行字符串，例如： a < b or c < d and e < f

每个符号都用空格间隔。

Output

 假链跳到０，真链跳到１，表达式序号从100开始排。

Sample Input

1
a < b or c < d and e < f

Sample Output

100(j<,a,b,1)
101(j,_,_,102)
102(j<,c,d,104)
103(j,_,_,0)
104(j<,e,f,100)
105(j,_,_,103)

Hint

Source

#include<bits/stdc++.h>
#include<vector>
#include<string>
using namespace std;
vector<string>to;
int main()
{
    int n ;
    scanf("%d",&n);
    getchar();
    while(n--)
    {
        to.clear();
        string s;
        getline(cin,s);
        s+=" #";
        string str;
        stringstream ss(s);
        int total  = 100;
        int True = 1;
        int False = 0;
        while(ss >> str)
        {
            if(str=="or")
            {
                cout<<total<<"(j" << to[1] << "," << to[0] << "," << to[2] <<"," << True<< ")" << endl;
                True = total;
                total++;
                printf("%d(j,_,_,%d)\n",total,total+1);
                total++;
                to.clear();
            }
            else if(str=="and")
            {
                cout<<total<<"(j" << to[1] << "," << to[0] << "," << to[2] << "," << total+2 << ")" << endl;
                total++;
                printf("%d(j,_,_,%d)\n",total,False);
                False = total;
                total++;
                to.clear();
            }
            else if(str=="#")
            {
                cout<<total<<"(j" << to[1] << "," << to[0] << "," << to[2] << "," << True << ")" << endl;
                total++;
                printf("%d(j,_,_,%d)\n",total,False);
                total++;
                to.clear();
            }
            else
            {
                to.push_back(str);
            }
       }
    }
    return 0;
}