2018 Multi-University Training Contest 4.Problem K. Expression in Memories(字符串模拟)

Problem K. Expression in Memories

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 4072    Accepted Submission(s): 864
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

解题思路: 

注意在类似 +0? 的情况下, ? 须被替换为 + 或 * ,其余情况直接将 ? 替换为非零数字就好。替换完成后判断一下是否合法。

#include <bits/stdc++.h>
using namespace std;

int T;
string res;

//判断是否为字符
bool judge(char c) {
    return (c=='+'||c=='*')?true:false;
}

int main()
{
    ios :: sync_with_stdio(0);
    cin >> T;
    while (T --)
    {
        cin >> res;
        int flag = 1,len=res.size();
        //改换策略
        for (int i = 1; i < len; ++ i)
            if (res[i] == '?' && res[i - 1] == '0' && (i == 1 || judge(res[i-2])))
                res[i] = '+';
        for (int i = 0; i < len; ++ i)
            if (res[i] == '?') res[i] = '1';
        //是否合法
        if (judge(res[0])||judge(res[len-1])) flag = 0;
        for (int i = 0; i < len - 1; ++ i)
            if ((judge(res[i]) && judge(res[i+1]))||((i == 0 ||judge(res[i-1])) && (res[i] == '0') && !judge(res[i+1])))
                flag = 0;
        //结果输出
        if (flag) cout << res << endl;
        else cout << "IMPOSSIBLE" << endl;
    }
    return 0;
}

猜你喜欢

转载自blog.csdn.net/XxxxxM1/article/details/81354067