hdu Problem K. Expression in Memories

              Problem K. Expression in Memories

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 262144/262144K (Java/Other)

Total Submission(s) : 1   Accepted Submission(s) : 1

Special Judge

Problem Description

Kazari remembered that she had an expression $s_0$ 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 $s_0$ 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 $s_0$ 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 $s$ $(1 \le |s| \le 500, \sum {|s|} \le 10 ^ 5)$. It is guaranteed that each character of $s$ will be in 0123456789+*? .

Output

For each test case, print a string $s_0$ 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

Source

2018 Multi-University Training Contest 4

这道题主要是模拟，考虑所有情况即可，需要注意有+0?,需要将?改成1，首位和最后一位不能是+/*，

1+0？，01，++/+*/**/*+，这些情况都不对

#include<bits/stdc++.h>
using namespace std;
char a[101000];
int len;
int pd()
{
    int i;
    if(a[0]=='+'||a[0]=='*')
        return 0;
    if(a[len-1]=='+'||a[len-1]=='*')
     return 0;
    for(i=0;i<len;i++)
    {
        if(a[i]=='+'||a[i]=='*')
        {
                if(a[i+1]=='+'||a[i+1]=='*')
                {
                    return 0;
                }
        }
        if(a[i]=='0')
        {
            if(a[i-1]=='+'||a[i-1]=='*')
            {
                if(a[i+1]>='0'&&a[i+1]<='9')
                {
                    return 0;
                }
            }
        }
        if(i==0)
        {
            if(a[i]=='0')
            {
                if(a[i+1]>='0'&&a[i+1]<='9')
                {
                    return 0;
                }
            }
        }
    }
    return 1;
}
int main()
{
    int t;
    scanf("%d",&t);
    getchar();
    while(t--)
    {
        memset(a,0,sizeof(a));
        scanf("%s",a);
        len= strlen (a);
        if(a[0]=='0')
        {
            if(a[1]=='?')
            {
                a[1]='+';
            }
        }
        for(int i=1;i<len-1;i++)
        {
            if(a[i]=='0')
            {
                if(a[i-1]=='+'||a[i-1]=='*')
                {
                    if(a[i+1]=='?')
                    {
                        a[i+1]='+';
                    }
                }
            }
        }
        for(int i=0;i<len;i++)
        {
            if(a[i]=='?')
            {
                a[i]='1';
            }
        }
        int k = pd();
        if(k==0)
        {
            printf("IMPOSSIBLE\n");
        }
        else
        {
            puts(a);
        }

    }
    return 0;
}