Experimental Educational Round: VolBIT Formulas Blitz C

最新推荐文章于 2024-03-17 23:03:12 发布

转载最新推荐文章于 2024-03-17 23:03:12 发布 · 94 阅读

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原文链接：http://www.cnblogs.com/yinghualuowu/p/5208367.html

探讨了在税局新大楼中，如何使用仅由数字7和8组成的幸运数字为办公室门牌号，以确保每个办公室都有独特的编号。通过算法计算了在门牌号长度限制下，最多可以有多少个办公室。

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.

Examples

input

output

6
1位数的，7 8
2位数的，77 78 87 88
3位数的，777 778 787 788 877 878 887 888
.....

#include<stdio.h>
//#include<bits/stdc++.h>
#include<string.h>
#include<iostream>
#include<math.h>
#include<sstream>
#include<set>
#include<queue>
#include<map>
#include<vector>
#include<algorithm>
#include<limits.h>
#define inf 0x7fffffff
#define INF 0x7fffffffffffffff
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define ULL unsigned long long
using namespace std;
LL  powl(LL x, LL n)
{
    LL pw = 1;
    while (n > 0)
    {
        if (n & 1)        // n & 1 等价于 (n % 2) == 1
            pw *= x;
        x *= x;
        n >>= 1;        // n >>= 1 等价于 n /= 2
    }
    return pw;
}
int main()
{
    LL n;
    LL num;
    cin>>n;
    cout<<powl(2,n+1)-2<<endl;
   // printf("%.20f\n",n*powl(1.000000011,num));
    return 0;
}

转载于:https://www.cnblogs.com/yinghualuowu/p/5208367.html