1079 延迟的回文数 (20 分)

“天真”的我以为就算输入的是回文数也要变，最后就直接在 do循环 上方加了一个回文判断，代码就成这样了~~~

#include<iostream>
#include<algorithm>
using namespace std;
int main(){
	int k=0,f=1;
	string s;
	cin>>s;
	for(int i=0;i<s.length()/2;i++)if(s[i]!=s[s.length()-1-i])f=0;
    if(f!=1)
	do{
		string x=s,y=s,c;
		reverse(y.begin(),y.end());
		int jw=0;
		for(int i=s.length()-1;i>=0;i--){
			int m=x[i]-'0'+y[i]-'0'+jw;
			c+=m%10+'0';
			jw=m/10;
		}
		if(jw!=0)c+=jw+'0';
		reverse(c.begin(),c.end());
		int w=0;f=1;
		while(c[w]=='0')w++;
		s=c.substr(w);
		for(int i=0;i<s.length()/2;i++)if(s[i]!=s[s.length()-1-i])f=0;
		cout<<x<<" + "<<y<<" = "<<s<<endl;
	}while(f==0&&++k<10);
	if(f==1)cout<<s<<" is a palindromic number.";
	else if(k==10)cout<<"Not found in 10 iterations.";
    return 0;
}