Digital Folding

Posted on 2026-02-06 20:20  贾普鲁登  阅读(0)  评论(0)    收藏  举报 
G-Digital Folding_2026牛客寒假算法基础集训营1

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 inline long long renum(string s,int l)
 4 {
 5     long long pi = 1,a = 0;
 6     for(int i = 0;i < l;i++)
 7     {
 8         a += pi * int(s[i] - '0');
 9         pi *= 10; 
10     }
11     return a;
12 }
13 int main()
14 {
15     int T,ln,rn;
16     string l,r,ans;
17     cin >> T;
18     while(T--)
19     {
20         l.clear();
21         r.clear();
22         ans.clear();
23         cin >> l >> r;
24         ln = l.size();
25         rn = r.size();
26         if(l == r)
27         {
28             cout << renum(l,ln) << endl;
29             continue;
30         }
31         if(renum(r,rn) == 1)
32         {
33             for(int i = 1;i <= rn - 1;i++) cout << "9";
34             cout << endl;
35             continue;
36         }
37         if(ln < rn)
38         {
39             l[0] = '1';
40             for(int i = 1;i < rn;i++) l[i] = '0';
41             ln = rn;
42         }
43         bool flag = 0;
44         //对于未分配内存的string 直接修改内容会导致内容存储不稳定 因用push_back
45         for(int i = 0;i < rn;i++)
46         {
47             if(flag == 1) ans.push_back('9');
48             else if(l[i] == r[i]) ans.push_back(l[i]);
49             else
50             {
51                 flag = 1;
52                 ans.push_back(char(r[i] - 1));
53             }
54         } 
55         cout << ((renum(ans,rn) > renum(r,rn)) ? renum(ans,rn) : renum(r,rn)) << endl;
56     }
57 }