1136 A Delayed Palindrome

题目描述
知识点： 大数加法，模拟
思路： 一个简单的大数加法 + 判断回文数进行模拟。

#include
#include
#include
using namespace std;
vector<int> add(vector<int> a,vector<int> b){
    vector<int> c;
    int t = 0;
    for(int i = a.size()-1;i >= 0;i--){
        t += a[i] + b[i];
        c.push_back(t % 10);
        t /= 10;
    }
    if(t) c.push_back(t);
    reverse(c.begin(),c.end());
    return c;
}
bool check(vector<int> a){
    for(int i = 0,j = a.size()-1;i < j;i++,j--){
        if(a[i] != a[j]) return false;
    }
    return true;
}
void print(vector<int> a){
    for(auto i : a) cout<<i;
}
int main(){
    string s;
    cin>>s;
    vector<int> a;
    for(auto i : s)
        a.push_back(i-'0');
    vector<int> c = a;
    int ite = 0;
    while(!check(c) && ite < 10){
        vector<int> b = c;
        a = c;
        reverse(b.begin(),b.end());
        c = add(a,b);
        ite++;
        print(a);
        cout<<" + ";
        print(b);
        cout<<" = ";
        print(c);
        cout<<endl;
        
    }
    if(ite >= 10) cout<<"Not found in 10 iterations.";
    else {
        print(c);
        cout<<" is a palindromic number.";
    }
    return 0;
}
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