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  • バイナリハックイージー / Unhappy Hacking (ABC Edit) (stack)

    题目链接:http://abc043.contest.atcoder.jp/tasks/abc043_b

    Time limit : 2sec / Memory limit : 256MB

    Score : 200 points

    Problem Statement

    Sig has built his own keyboard. Designed for ultimate simplicity, this keyboard only has 3 keys on it: the 0 key, the 1 key and the backspace key.

    To begin with, he is using a plain text editor with this keyboard. This editor always displays one string (possibly empty). Just after the editor is launched, this string is empty. When each key on the keyboard is pressed, the following changes occur to the string:

    • The 0 key: a letter 0 will be inserted to the right of the string.
    • The 1 key: a letter 1 will be inserted to the right of the string.
    • The backspace key: if the string is empty, nothing happens. Otherwise, the rightmost letter of the string is deleted.

    Sig has launched the editor, and pressed these keys several times. You are given a string s, which is a record of his keystrokes in order. In this string, the letter 0 stands for the 0 key, the letter 1 stands for the 1 key and the letter B stands for the backspace key. What string is displayed in the editor now?

    Constraints

    • 1≦|s|≦10 (|s| denotes the length of s)
    • s consists of the letters 01 and B.
    • The correct answer is not an empty string.

    Input

    The input is given from Standard Input in the following format:

    s
    

    Output

    Print the string displayed in the editor in the end.


    Sample Input 1

    Copy
    01B0
    

    Sample Output 1

    Copy
    00
    

    Each time the key is pressed, the string in the editor will change as follows: 001000.


    Sample Input 2

    Copy
    0BB1
    

    Sample Output 2

    Copy
    1
    

    Each time the key is pressed, the string in the editor will change as follows: 0(empty)(empty)1.

    题解:栈 我也不知道为啥有三组数据过不了 先放这

      大晚上来补这道题 后来才发现自己的思路错了 后面输出的顺序完全反了 (栈的特性 最后只需要改改顺序就行了)思路错了居然还过了一大半的数据 有毒 ...

     1 #include <iostream>
     2 #include <algorithm>
     3 #include <cstring>
     4 #include <cstdio>
     5 #include <vector>
     6 #include <cstdlib>
     7 #include <iomanip>
     8 #include <cmath>
     9 #include <ctime>
    10 #include <map>
    11 #include <set>
    12 #include <queue>
    13 #include <stack>
    14 using namespace std;
    15 #define lowbit(x) (x&(-x))
    16 #define max(x,y) (x>y?x:y)
    17 #define min(x,y) (x<y?x:y)
    18 #define MAX 100000000000000000
    19 #define MOD 1000000007
    20 #define pi acos(-1.0)
    21 #define ei exp(1)
    22 #define PI 3.141592653589793238462
    23 #define INF 0x3f3f3f3f3f
    24 #define mem(a) (memset(a,0,sizeof(a)))
    25 typedef long long ll;
    26 ll gcd(ll a,ll b){
    27     return b?gcd(b,a%b):a;
    28 }
    29 bool cmp(int x,int y)
    30 {
    31     return x>y;
    32 }
    33 const int N=10005;
    34 const int mod=1e9+7;
    35 int main()
    36 {
    37     std::ios::sync_with_stdio(false);
    38     string a;
    39     cin>>a;
    40     int len=a.length();
    41     stack <char> q ;
    42     for(int i=0;i<len;i++){
    43         if(a[i]=='1')  q.push('1');
    44         else if(a[i]=='0')  q.push('0');
    45         else if(a[i]=='B'){
    46             if(q.size()>0)
    47                 q.pop();
    48         }
    49     }
    50     char b[20];
    51     int t=0,i;
    52     while(q.size()>0){
    53         b[t++]=q.top();
    54         q.pop();
    55     }
    56     for(i=t-1;i>=0;i--)
    57         cout<<b[i];
    58     cout<<endl;
    59     return 0;
    60 }
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  • 原文地址:https://www.cnblogs.com/wydxry/p/7287232.html
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