CODEFORCES630C- Lucky Numbers(水题)

本文介绍了一个算法,用于生成由数字7和8组成的幸运数字,并计算在限定位数内可能生成的最大数量的独特幸运数字。该算法适用于税务办公室新大楼办公室编号的需求。
 Lucky Numbers
Time Limit:500MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

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
2
Output

6


#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
using namespace std;
__int64 a[60];
void dabiao()
{
	int i,j;
	a[1]=2;
	for(i=2;i<=60;i++)
	{
		long long  temp=1;
		for(j=1;j<=i;j++)
		{
			temp*=2;
		}
		a[i]=a[i-1]+temp;
	}
}
int main()
{
	dabiao();
	int n;
	while(scanf("%d",&n)!=EOF)
	{
		printf("%I64d\n",a[n]);
	} 
} 


### Codeforces Problem 976C Solution in Python For solving problem 976C on Codeforces using Python, efficiency becomes a critical factor due to strict time limits aimed at distinguishing between efficient and less efficient solutions[^1]. Given these constraints, it is advisable to focus on optimizing algorithms and choosing appropriate data structures. The provided code snippet offers insight into handling string manipulation problems efficiently by customizing comparison logic for sorting elements based on specific criteria[^2]. However, for addressing problem 976C specifically, which involves determining the winner (&#39;A&#39; or &#39;B&#39;) based on frequency counts within given inputs, one can adapt similar principles of optimization but tailored towards counting occurrences directly as shown below: ```python from collections import Counter def determine_winner(): for _ in range(int(input())): count_map = Counter(input().strip()) result = "A" if count_map[&#39;A&#39;] > count_map[&#39;B&#39;] else "B" print(result) determine_winner() ``` This approach leverages `Counter` from Python’s built-in `collections` module to quickly tally up instances of &#39;A&#39; versus &#39;B&#39;. By iterating over multiple test cases through a loop defined by user input, this method ensures that comparisons are made accurately while maintaining performance standards required under tight computational resources[^3]. To further enhance execution speed when working with Python, consider submitting codes via platforms like PyPy instead of traditional interpreters whenever possible since they offer better runtime efficiencies especially important during competitive programming contests where milliseconds matter significantly.
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