题目：

Description
Chilly Willy loves playing with numbers. He only knows prime numbers that are digits yet. These numbers are 2, 3, 5 and 7. But Willy grew rather bored of such numbers, so he came up with a few games that were connected with them.
Chilly Willy wants to find the minimum number of length n, such that it is simultaneously divisible by all numbers Willy already knows (2, 3, 5 and 7). Help him with that.
A number's length is the number of digits in its decimal representation without leading zeros.
Input
A single input line contains a single integer n (1 ≤ n ≤ 105).
Output
Print a single integer — the answer to the problem without leading zeroes, or "-1" (without the quotes), if the number that meet the problem condition does not exist.
Sample Input
Input
1
Output
-1
Input
5
Output
10080

给定一个数n，要你找出一个数，这个数有n位且是最小的能被2，3，5，7都整除的数。
n的范围很大，因此不能直接去做。
事实上，能被2，3，5，7都整除的数就是能被210整除的数（最小公倍数）。通过观察，我们可以发现只需考虑后3位数即可，最后一位一定是0，所以只需考虑倒数第三位和倒数第二位即可。通过运算我们可以找到规律：从n = 4开始，6个为一个循环。

参考代码：

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

char str[7][3] = {"05", "08", "17", "02", "20", "11"};

void print(const int n) {
    if (n == 1 || n == 2) cout << "-1" << endl;
    else if (n == 3) cout << "210" << endl;
    else {
        int temp = (n - 4) % 6;
        printf("1");
        for (int i = 0;i < n - 4;++i) {//剩余位补0;
            printf("0");
        }
        printf("%s0\n", str[temp]);
    }   
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    int n;
    cin >> n;
    print(n);
    return 0;
}

其实我开始也不知道这个题的规律，是看了网上大神们的博客才知道这个规律的，还是大神们厉害。