630C-

最新推荐文章于 2024-11-14 10:42:57 发布

原创最新推荐文章于 2024-11-14 10:42:57 发布 · 244 阅读

0 ·

CC 4.0 BY-SA版权

编程语言同时被 3 个专栏收录

307 篇文章

订阅专栏

==数学and思维==

131 篇文章

订阅专栏

codeforces

76 篇文章

订阅专栏

本文介绍了一种算法，用于计算在限定门牌号长度的情况下，由仅包含数字7和8组成的独特幸运数字作为门牌号的办公室最大数量。通过递推公式预计算不同长度下可能的最大办公室数量。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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.

Example

Input

Output

房间号是幸运数字的房间一共是多少间（每个房间房间号都不同），注意幸运号数字是指组成房间号的数字只能是7或8，只有这两种数字种类

#include<stdio.h>
long long num[60];
long long time(int i){
	int j;
	long long n=1;
	for(j=1;j<=i;j++){
		n*=2;
	}
	return n;
}
/*long long p(int n,int m)
{
	int t=1;
	while(m)
	{
		if(m%2==1)
		t=t*n;
		m=m/2;
	    n=n*n;
	}
	return t;
}*/
int main(){
	int n,i;
	num[0]=0;
	for(i=1;i<=55;i++){
		num[i]=num[i-1]+time(i);
	}
	while(scanf("%d",&n)!=EOF){
	printf("%lld\n",num[n]);
	}
	return 0;
}

#include<cstdio>
using namespace std;
long long p(long long n,long long m)
{
	long long t=1;
	while(m>0)
	{
		if(m%2==1)
		t=t*n;
		n=n*n;
		m=m/2;
	 } 
	 return t;
}
int main()
{
long long n;
long long a[1000];
a[0]=0;
for(int i=1;i<=1000;i++)
a[i]=a[i-1]+p(2,i);
while(~scanf("%lld",&n))
{
	printf("%lld\n",a[n]);
}
return 0;	
}