PTA乙级（1079 延迟的回文数 (20分)）

1079 延迟的回文数 (20分)

https://pintia.cn/problem-sets/994805260223102976/problems/994805261754023936

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <string>
 4 #include <iostream>
 5 #include <algorithm>
 6 #include <cmath>
 7 #include <cstdlib>
 8 using namespace std;
 9 bool ispalindrome(string a);
10 string add(string b,string c);
11 int main()
12 {
13     string a,b,c;
14     bool flag=false;
15     cin>>a;
16     if(ispalindrome(a)) {cout<<a<<" is a palindromic number.\n"; return 0;}
17     int cnt=10;
18     while(cnt>0)
19     {
20         b=a;
21         reverse(b.begin(),b.end());
22         c=add(a,b);
23         cout<<a<<" + "<<b<<" = "<<c<<endl;
24         if(ispalindrome(c)) {cout<<c<<" is a palindromic number.\n"; flag=true; break;}
25         a=c;
26         cnt--;
27     }
28     if(!flag) cout<<"Not found in 10 iterations.\n";
29     return 0;
30 }
31 bool ispalindrome(string a)
32 {
33     int len=a.size();
34     for(int i=0;i<len/2;i++)
35     {
36         if(a[i]!=a[len-1-i]) return 0;
37     }
38     return 1;
39 }
40 string add(string a,string b)
41 {
42     string d;
43     int adv=0,bac=0;
44     int len=a.size();
45     for(int i=len-1;i>=0;i--)
46     {
47         bac=(a[i]-'0')+(b[i]-'0');
48         d+=(bac+adv)%10+'0';
49         adv=(bac+adv)/10;
50     }
51     if(adv) d+=(adv)+'0';
52     reverse(d.begin(),d.end());
53     return d;
54 }