Codeforces 934.B A Prosperous Lot
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 1018, such that there are exactly k loops in the decimal representation of n, or determine that such n does not exist.
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.
![](http://codeforces.com/predownloaded/c6/6b/c66bf8ecc5d9ef3b7001a32bf05d4a0d784f311b.png)
The first and only line contains an integer k (1 ≤ k ≤ 106) — the desired number of loops.
Output an integer — if no such n exists, output -1; otherwise output any such n. In the latter case, your output should be a positivedecimal integer not exceeding 1018.
2
462
6
8080
题目大意:每个数字都有其封闭空间的数量,现在要求给出一个不超过10^18的正数,使得这个数字所有的数位上的数字的封闭区间的个数和等于k.
分析:水题啊......如果k > 36就无解了. 不然肯定先填8,n为奇数就填一个4.
#include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; typedef long long ll; int n; int main() { scanf("%d",&n); if (n > 36) puts("-1"); else { int num1 = n / 2; int num2 = n - num1 * 2; for (int i = 1; i <= num1; i++) printf("8"); for (int i = 1; i <= num2; i++) printf("4"); } return 0; }