zoukankan      html  css  js  c++  java
  • 杭电多校第四场 Problem K. Expression in Memories 思维模拟

    Problem K. Expression in Memories

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
    Total Submission(s): 0    Accepted Submission(s): 0
    Special Judge


    Problem Description
    Kazari remembered that she had an expression s0 before.
    Definition of expression is given below in Backus–Naur form.
    <expression> ::= <number> | <expression> <operator> <number>
    <operator> ::= "+" | "*"
    <number> ::= "0" | <non-zero-digit> <digits>
    <digits> ::= "" | <digits> <digit>
    <digit> ::= "0" | <non-zero-digit>
    <non-zero-digit> ::= "1" | "2" | "3" | "4" | "5" | "6" | "7" | "8" | "9"
    For example, `1*1+1`, `0+8+17` are valid expressions, while +1+1, +1*+1, 01+001 are not.
    Though s0 has been lost in the past few years, it is still in her memories. 
    She remembers several corresponding characters while others are represented as question marks.
    Could you help Kazari to find a possible valid expression s0 according to her memories, represented as s, by replacing each question mark in s with a character in 0123456789+* ?
     
    Input
    The first line of the input contains an integer T denoting the number of test cases.
    Each test case consists of one line with a string s (1|s|500,|s|105).
    It is guaranteed that each character of s will be in 0123456789+*? .
     
    Output
    For each test case, print a string s0 representing a possible valid expression.
    If there are multiple answers, print any of them.
    If it is impossible to find such an expression, print IMPOSSIBLE.
     
    Sample Input
    5 ????? 0+0+0 ?+*?? ?0+?0 ?0+0?
     
    Sample Output
    11111 0+0+0 IMPOSSIBLE 10+10 IMPOSSIBLE

    先考虑运算符不行的情况再考虑前导0的情况 

    AC代码
    #include <map>
    #include <set>
    #include <stack>
    #include <cmath>
    #include <queue>
    #include <cstdio>
    #include <vector>
    #include <string>
    #include <cstring>
    #include <iomanip>
    #include <iostream>
    #include <algorithm>
    #define ls (r<<1)
    #define rs (r<<1|1)
    #define debug(a) cout << #a << " " << a << endl
    using namespace std;
    typedef long long ll;
    const ll maxn = 1e5 + 10;
    const ll mod = 1e9 + 7;
    string s;
    vector<string> e;
    bool iso( char c ) {
        if( c == '+' || c == '*' ) {
            return true;
        }
        return false;
    }
    int main() {
        ll T;
        cin >> T;
        while( T -- ) {
            cin >> s;
            bool flag = true;
            for( ll i = 0; i < s.length(); i ++ ) {
                if( ( i == 0 || i == s.length()-1 ) && iso(s[i]) ) {
                    flag = false;
                    break;
                }
                if( i < s.length()-1 ) {
                    if( iso(s[i]) && iso(s[i+1]) ) {
                        flag = false;
                        break;
                    }
                }
                if( s[i] == '?' ) {
                    if( s[i-1] == '0' && ( i-2 < 0 || iso(s[i-2])  ) && !iso(s[i+1]) ) {
                        s[i] = '+';
                    } else {
                        s[i] = '1';
                    }
                }
            }
            //debug(s);
            e.clear();
            string t = "";
            for( ll i = 0; i < s.length(); i ++ ) {
                if( iso(s[i]) || i == s.length()-1 ) {
                    if( i == s.length()-1 ) {
                        t += s[i];
                    }
                    e.push_back(t);
                    t = "";
                } else {
                    t += s[i];
                }
            }
            for( ll i = 0; i < e.size(); i ++ ) {
                //cout << e[i] << endl;
                if( e[i][0] == '0' && e[i].length() > 1 ) {
                    flag = false;
                    break;
                }
            }
            if( flag ) {
                cout << s << endl;
            } else {
                cout << "IMPOSSIBLE" << endl;
            }
        }
        return 0 ;
    }
    

      

    彼时当年少,莫负好时光。
  • 相关阅读:
    LoadRunner的Capture Level说明
    LoadRunner Click&script 录制Tips
    LoadRunner虚拟用户协议脚本语言矩阵表
    LoadRunner 测试 AJAX
    如何创建自定义性能计数器
    8个批量样本数据生成工具
    JDBC性能优化
    使用Servlet为LoadRunner提供全局连续唯一数
    LoadRunner11测试Weblogic的问题
    如何让Fiddler捕获并记录HTTPS包?
  • 原文地址:https://www.cnblogs.com/l609929321/p/9402017.html
Copyright © 2011-2022 走看看