计算机挑战赛1-优快云博客

从大于等于N的正整数里找到一个最小的数M，使之满足:M和M的逆序数(如1230的逆序数为321)的乘积为一个[100000000,200000000]区间内的值。输入说明:起始数输出说明:找到的第一个符合条件的数，如果没有符合条件的数，输出F。输入字N;样例:123456输出样例:124100

代码：

C++：

#include<iostream>
#include<string>
using namespace std;
int main() {
	long long n; cin >> n;
	bool flag = true;
	for (auto i = n; i <= 200000000; i++) {
		string s;
		s = to_string(i);
		string s1 = "";
		for (auto j = 0; j < s.length(); j++) {
			s1 += s[s.length() - 1 - j];
		}
		int num = stoi(s1);
		if (i * num >= 100000000 && i * num <= 200000000) {
			cout << i << endl;
			flag = false;
			break;
		}
	}
	if (flag == true) {
		cout << "F" << endl;
	}
	return 0;
}

Java:

package com.my.gududu;

import java.util.*;

public class Main {
    public static void main(String[] args) {
        boolean flag = true;
        Scanner input = new Scanner(System.in);
        int n = input.nextInt();
        for (int i = n; i <= 200000000; i++) {
            StringBuilder s = new StringBuilder(String.valueOf(i));
            String s1 = "";
            for (int j = 0; j < s.length(); j++) {
                s1 += s.charAt(s.length() - 1 - j);
            }
            int num = Integer.valueOf(s1);
            if (i * num >= 100000000 && i * num <= 200000000) {
                System.out.println(i);
                flag = false;
                break;
            }
        }
        if (flag == true) {
            System.out.println("F");
        }
    }
}

python:

n = int(input())
flag = True
for i in range(n, 200000001, 1):
    s1 = ''
    s = str(i)
    for j in range(len(s)):
        s1 += s[len(s) - 1 - j]
    num = int(s1)
    if i * num >= 100000000 and i * num <= 200000000:
        flag = False
        print(i)
        break
if flag == True:
    print("F")