题解 | #回文对称数#

知识点：

循环控制：循环控制

解法一：字符串逆序

解法二：数字逆序

#include <iostream>
#include <algorithm>
using namespace std;

int main() {
    int n;

    cin >> n;

    for (int i = 1; i <= n; i++) {
        int sum = 0;
        int num = i;

        while (num > 0) {
            sum = sum * 10 + num % 10;
            num /= 10;
        }

        if (i == sum) {
            cout << i << endl;
        }
    }

    return 0;
}

// 解法一：字符串
/*
int main() {
    int n;
    string str_o;
    string str_r;

    cin >> n;

    for (int i = 1; i <= n; i++) {
        str_r = str_o = to_string(i);
        reverse(str_r.begin(), str_r.end());
        if (str_o == str_r) {
            cout << i << endl;
        }
    }

    return 0;
}
*/