任意门：http://acm.hdu.edu.cn/showproblem.php?pid=6342

Problem K. Expression in Memories

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 2150    Accepted Submission(s): 772
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

Source

2018 Multi-University Training Contest 4

题意概括：

给一串表达式（可能完整可能不完整），表达式只含有 ‘+’ 和 ‘ * ’ 两种运算，数字为 0~9；

如果不完整（含‘ ? ‘)， 则补充完整。

若表达式本身非法或者无法补充成为一个合法表达式，则输出“IMPOSSIBLE”

解题思路：

很明显 “ ？” 只有在 0？的情况下需要变成 ‘+’ 或者‘*’; 其他情况都把 “ ？”变成 非0的数字即可。

判断表达式是否合法： 是否出现 0111 或者 ++ 或者 *+ 这类的情况即可。

AC code：

 1 #include<cstdio>
 2 #include<algorithm>
 3 #include<iostream>
 4 #include<cstring>
 5 #include<vector>
 6 #include<queue>
 7 #include<cmath>
 8 #include<set>
 9 #define INF 0x3f3f3f3f
10 #define LL long long
11 using namespace std;
12 const LL MOD = 1e9+7;
13 const int MAXN = 555;
14 char str[MAXN];
15 //char ans[MAXN];
16
17 int main()
18 {
19     int T_case;
20     scanf("%d", &T_case);
21     while(T_case--){
22         scanf("%s", str);
23         int len = strlen(str);
24         bool flag = true;
25
26         for(int i = 0; i < len; i++){
27             if(str[i] == ‘+‘ || str[i] == ‘*‘){
28                 if(i == 0 || i == len-1){
29                     flag = false;
30                     break;
31                 }
32                 else if(str[i+1] == ‘+‘ || str[i+1] == ‘*‘){
33                     flag = false;
34                     break;
35                 }
36             }
37             else if(str[i] == ‘0‘){
38                 if(i == 0 || str[i-1] == ‘+‘ || str[i-1] == ‘*‘){
39                     if(i < len-1 && str[i+1] >= ‘0‘ && str[i+1] <= ‘9‘){
40                         flag = false;
41                         break;
42                     }
43                     else if(i < len-1 && str[i+1] == ‘?‘){
44                         str[i+1] = ‘+‘;
45                     }
46                 }
47             }
48             else if(str[i] == ‘?‘){
49                 str[i] = ‘1‘;
50             }
51         }
52
53         if(flag) printf("%s\n", str);
54         else puts("IMPOSSIBLE");
55     }
56     return 0;
57 }

原文地址：https://www.cnblogs.com/ymzjj/p/10335900.html