Codeforces Beta Round #8A Train and Peter (string的运用)

Posted 2020-08-02 AC_Dreameng

Train and Peter

Time Limit: 1000ms

Memory Limit: 65536KB

This problem will be judged on CodeForces. Original ID: 8A
64-bit integer IO format: %I64d      Java class name: (Any)

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Peter likes to travel by train. He likes it so much that on the train he falls asleep.

Once in summer Peter was going by train from city A to city B, and as usual, was sleeping. Then he woke up, started to look through the window and noticed that every railway station has a flag of a particular colour.

The boy started to memorize the order of the flags‘ colours that he had seen. But soon he fell asleep again. Unfortunately, he didn‘t sleep long, he woke up and went on memorizing the colours. Then he fell asleep again, and that time he slept till the end of the journey.

At the station he told his parents about what he was doing, and wrote two sequences of the colours that he had seen before and after his sleep, respectively.

Peter‘s parents know that their son likes to fantasize. They give you the list of the flags‘ colours at the stations that the train passes sequentially on the way from A to B, and ask you to find out if Peter could see those sequences on the way from A to B, or from B to A. Remember, please, that Peter had two periods of wakefulness.

Peter‘s parents put lowercase Latin letters for colours. The same letter stands for the same colour, different letters — for different colours.

Input

The input data contains three lines. The first line contains a non-empty string, whose length does not exceed 105, the string consists of lowercase Latin letters — the flags‘ colours at the stations on the way from A to B. On the way from B to A the train passes the same stations, but in reverse order.

The second line contains the sequence, written by Peter during the first period of wakefulness. The third line contains the sequence, written during the second period of wakefulness. Both sequences are non-empty, consist of lowercase Latin letters, and the length of each does not exceed 100 letters. Each of the sequences is written in chronological order.

Output

Output one of the four words without inverted commas:

• ?forward? — if Peter could see such sequences only on the way from A to B;
• ?backward? — if Peter could see such sequences on the way from B to A;
• ?both? — if Peter could see such sequences both on the way from A to B, and on the way from B to A;
• ?fantasy? — if Peter could not see such sequences.

Sample Input

Input

atob
a
b

Output

forward

Input

aaacaaa
aca
aa

Output

both

Hint

It is assumed that the train moves all the time, so one flag cannot be seen twice. There are no flags at stations A and B.

 题意:输入三个字符串,s,a,b,s为源串,a,b为目标串,问是否可以依次得到a串和b串,或相反.

此题本来可以用字符串模拟的,但是C++有强大的string类,怎么能不用呢?

看了q神的代码,"佩服"两字油然而生!

顺便补一下相关的函数:

查找(find)
语法:
size_type find( const basic_string &str, size_type index );
size_type find( const char *str, size_type index );
size_type find( const char *str, size_type index, size_type length );
size_type find( char ch, size_type index );

find()函数:

返回str在字符串中第一次出现的位置（从index开始查找）。如果没找到则返回string::npos,
返回str在字符串中第一次出现的位置（从index开始查找，长度为length）。如果没找到就返回string::npos,
返回字符ch在字符串中第一次出现的位置（从index开始查找）。如果没找到就返回string::npos
例如:

string str1( "Alpha Beta Gamma Delta" );
unsigned int loc = str1.find( "Omega", 0 );
if( loc != string::npos )
    cout << "Found Omega at " << loc << endl;
else
    cout << "Didn't find Omega" << endl;//print this

AC代码:

#include <iostream>
#include <string>
#include <algorithm>
using namespace std;
int main()
{
    string s,a,b;
    while(cin>>s>>a>>b)
    {
        bool isok[2]= {0,0};
        for(int i=0; i<2; i++)
        {
            int idx=s.find(a);
            if(idx!=string::npos)
            {
                idx=s.find(b,idx+a.size());
                if(idx!=string::npos)
                    isok[i]=1;
            }
            reverse(s.begin(),s.end());
        }
        if(isok[0]&&isok[1])
            cout<<"both"<<endl;
        else if(isok[0])
            cout<<"forward"<<endl;
        else if(isok[1])
            cout<<"backward"<<endl;
        else
            cout<<"fantasy"<<endl;
    }
    return 0;
}