Expression in Memories

2019年11月6日 169次阅读

Expression in Memories

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 283    Accepted Submission(s): 104
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

 

 

 

思路：+0？这种情况？一定是+或者*，其他情况的？是非0的数字

AC代码：

#include<bits/stdc++.h>
using namespace std;
const int maxn = 100010;
char s[maxn];
int len;
int judge()
{
    int i;
    if(s[0]=='*'||s[0]=='+')
    return 0;
    if(s[len-1]=='*'||s[len-1]=='+')
    return 0;
    for(i=0;i<len;i++)
    {
       if(s[i]=='*'||s[i]=='+')
       {
          if(s[i+1]=='*'||s[i+1]=='+')
          return 0;
       }
       if(s[i]=='0')
       {
           if(s[i-1]=='*'||s[i-1]=='+')
           {
              if(s[i+1]>='0'&&s[i+1]<='9')
              return 0;
           }
       }
       if(i==0)
       {
          if(s[i]=='0')
          {
             if(s[i+1]>='0'&&s[i+1]<='9')
             return 0;
          }
       }
    }
    return 1;
}
int main()
{
    int n;
    scanf("%d",&n);
    getchar();
    //101010   01010  +0?
    while(n--){
        memset(s,0,sizeof(s));
        scanf("%s",&s);
        len = strlen(s);
        if(s[0]=='0'){ //先判断第一个是否是0后面是否是？，这算是一种特殊情况了
            if(s[1]=='?')
               s[1]='+';
        }
        for(int i=1;i<len-1;i++){
            if(s[i]=='0'){
                if(s[i-1]=='+' || s[i-1]=='*'){
                    if(s[i+1]=='?')
                      s[i+1]='+';
                }
            }
        }
        for(int i=0;i<len;i++){
            if(s[i]=='?')
            s[i]='1';
        }
        //puts(s);
        int ans = judge();
        if(ans==0){
            printf("IMPOSSIBLE\n");
        }
        else{
            puts(s);
        }
    }
    return 0;
}

 

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