Codeforces Round #462 (Div. 2) B. A Prosperous Lot_apart from nian there is a daemon named sui which -优快云博客

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本文介绍了一个有趣的数学问题，即如何找出一个不超过1e18的正整数，该整数的十进制表示中恰好包含k个封闭图形（环），如数字8有两个环，0有一个环。文章提供了一种解决方案，并附带了C++代码实现。

B. A Prosperous Lot

time limit per test1 second
memory limit per test256 megabytes

Problem Description

Apart from Nian, there is a daemon named Sui, which terrifies children and causes them to become sick. Parents give their children money wrapped in red packets and put them under the pillow, so that when Sui tries to approach them, it will be driven away by the fairies inside.

Big Banban is hesitating over the amount of money to give out. He considers loops to be lucky since it symbolizes unity and harmony.

He would like to find a positive integer n not greater than 1e18, such that there are exactly k loops in the decimal representation of n, or determine that such n does not exist.
这里写图片描述
A loop is a planar area enclosed by lines in the digits’ decimal representation written in Arabic numerals. For example, there is one loop in digit 4, two loops in 8 and no loops in 5. Refer to the figure below for all exact forms.

Input

The first and only line contains an integer k (1 ≤ k ≤ 1e6) — the desired number of loops.

Output

Output an integer — if no such n exists, output -1; otherwise output any such n. In the latter case, your output should be a positive decimal integer not exceeding 1e18.

Examples

input
2
output
462
input
6
output
8080

解题心得：

题意看了好一会儿，其实就是一个数字，可能会形成封闭图形，输入n，叫你输出一个数字不超过1e18，在这个数字中有n个封闭图形，如果不能得到一个这样的数字就输出-1；
其实只要找到8和0来组织就行了，只有8能形成两个封闭图形，0能形成一个封闭图形还是最小的个位数，数字大小不超过1e18，所以最多能形成36个封闭图形，如果n大于36直接输出-1；

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int main(){
    ll k;
    scanf("%lld",&k);
    if(k > 36){
        printf("-1");
        return 0;
    }
    int p = k/2;
    for(int i=0;i<p;i++)
        printf("8");
    if(k%2)
        printf("6");
    return 0;
}