Problem K. Expression in Memories

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本文介绍了一个有趣的问题：根据部分已知字符和问号组成的字符串，尝试还原一个有效的数学表达式。文章提供了一段代码实现，该算法通过替换问号来构造合法的数学表达式，并考虑了多种边界情况。

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

代码：

#include <bits/stdc++.h>
using namespace std;
char s[505];
int main(){
    int t,n,f;
    scanf("%d",&t);
    getchar();
    while(t--){
        memset(s,0,sizeof(s));
        f=0;
        gets(s);
        n=strlen(s);
        for(int i=1;i<n;i++){
            if((s[i]=='+'||s[i]=='*')&&((s[i-1]=='+'||s[i-1]=='*'))){f=1;break;}
        }
        if(s[0]=='+'||s[0]=='*'||s[n-1]=='+'||s[n-1]=='*')f=1;
        for(int i=0;i<n-1;i++){
            if((s[i]=='0')&&(i==0||(s[i-1]=='+'||s[i-1]=='*'))){
                if(s[i+1]!='+'&&s[i+1]!='*'&&s[i+1]!='?'){f=1;break;}
                if(s[i+1]=='?'){
                    if(i+1==n-1||(s[i+2]=='+'||s[i+2]=='*')){f=1;break;}
                    s[i+1]='+';
                    if(s[i+2]=='?')s[i+2]='1';
                }
            }
            if((s[i]=='?')&&(i==0||(s[i-1]=='+'||s[i-1]=='*')))s[i]='1';
            else if((s[i]=='?')&&((s[i-1]!='+'||s[i-1]!='*'))){
                s[i]='1';
            }
        }
        if(f){
            puts("IMPOSSIBLE");continue;
        }
        for(int i=0;i<n;i++)if(s[i]=='?')s[i]='1';
        cout<<s<<endl;
    }
    return 0;
}