Codeforces--630C--Lucky Numbers（快速幂）

最新推荐文章于 2019-04-04 20:02:07 发布

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最新推荐文章于 2019-04-04 20:02:07 发布

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分类专栏：数学

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本文介绍了一个计算特定长度内所有仅包含数字7和8的幸运数字总数的问题，并提供了一种使用快速幂运算的方法来求解该问题。通过递归分解，最终得出公式2^(n+1)-2，并给出了一段C++代码实现。

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C - Lucky Numbers

Crawling in process... Crawling failed Time Limit:500MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

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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 ≤ n ≤ 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

Input

Output

数字中只含有7或者8的成为Lucky Numbers，问n位以内的数字有多少个Lucky Numbers，总数应该就是2+2^2+....+2^n,数学公式变形之后就是2^(n+1)-2个，n+1可能比较大，所以快速幂解决

#include<iostream>
using namespace std;
#define LL long long
LL  p(LL x, LL n)
{
    LL pw = 1;
    while (n > 0)
    {
        if (n & 1)      
            pw *= x;
        x *= x;
        n >>= 1;
    }
    return pw;
}
int main()
{
    LL n;
    LL num;
    cin>>n;
    cout<<p(2,n+1)-2<<endl;
    return 0;
}