"
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 //代码效果参考：https://v.youku.com/v_show/id_XNjQwMDQxOTYzNg==.html
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
1 /*
2 2016年4月24日16:55:26
3 题意: 给定n 找到十进制数 长度为n的 能同时整除2 3 5 7的最小的数
4
5 要找到210的倍数 很明显只用考虑后三位 最后一位一定为0，
6 因此找规律可以得到6个数一个循环
7 ""05"", ""08"", ""17"", ""02"", ""20"", ""11""
8
9 明显个鬼啊
10 */
11
12
13
14 # include
15 # include
16 # include
17 # include
18 # include
19 # include
20 # include
21 # define LL long long
22 # define INF 0x3f3f3f3f
23 using namespace std;
24 const int N = 1e5 + 5;
2//代码效果参考：https://v.youku.com/v_show/id_XNjQwMDQxOTY3Ng==.html
526 char s【6】【3】 = {""05"", ""08"", ""17"", ""02"", ""20"", ""11""};
27
28 int main(void)
29 {
30 int n, tmp, i;
31 while (~scanf(""%d"", &n)){
32 if (n <= 2)
33 printf(""-1\n"");
34 else if (n == 3)
35 printf(""210\n"");
36 else {
37 printf(""1"");
38 for (i = 2; i <= n-3; i++)
39 printf(""0"");
40 tmp = (n - 4) % 6;
41 printf(""%s0\n"", s【tmp】);
42 }
43 }
44
45 return 0;
46 }
"
