「CF630C」Lucky Numbers

Portal

Portal2: Luogu

Description

The numbers of all offices in the new building of the Tax Office of IT City will have lucky numbers.

Lucky number is a number that consists of digits \(7\) and \(8\) only. Find the maximum number of offices in the new building of the Tax Office given that a door-plate can hold a number not longer than \(n\) digits.

Input

The only line of input contains one integer \(n (1 \le n \le 55)\) — the maximum length of a number that a door-plate can hold.

Output

Output one integer — the maximum number of offices, than can have unique lucky numbers not longer than \(n\) digits.

Sample Input

Sample Output

Solution

题目要我们构造\(1 \sim n\)位由\(7, 8\)的数的个数。我们先来找一找规律：

位数为\(1\)时：有\(7, 8\)，共\(2 \times 2 ^ 0 = 2\)种；

位数为\(2\)时：有\(77, 78, 87, 88\)，共\(2 \times 2 ^ 1 = 4\)种；

位数为\(3\)时：有\(777, 778, 787, 788, 877, 878, 887, 888\)共\(2 \times 2 ^ 2 = 8\)种；

\(\cdots \cdots\)

所以，位数是\(n\)的总个数是\(2 \times 2 ^ {n - 1}\)；

那么位数为\(1 \sim n\)的总个数为

\[\begin{aligned} \sum^{n}_{i = 1}{2 \times 2 ^ {i - 1}} & = 2 \times \sum^{n}_{i = 1}{2 ^ {i - 1}} \\\\ & = 2 \times (2 ^ {n} - 2) \\\\ & = 2 ^ {n + 1} - 2\end{aligned}\]

于是就解决了。

Code

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>

using namespace std;

typedef long long LL;
LL n;
inline LL power(LL x, LL y) {//求x的y次方
    LL ret = 1;
    for (LL i = 1; i <= y; i++)
        ret *= x;
    return ret;
}
int main() {
    scanf("%lld", &n);
    printf("%lld\n", power(2, n + 1) - 2);//推出来的公式
    return 0;
}