Problem--82A--Codeforces--Double Cola

最新推荐文章于 2020-08-01 16:17:08 发布

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最新推荐文章于 2020-08-01 16:17:08 发布

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本文探讨了一个有趣的队列问题，涉及五个人物购买并饮用DoubleCola后人数翻倍的过程，通过编程找出第n个人是谁。提供了两种不同的C语言实现方法。

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Double Cola
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output

Sheldon, Leonard, Penny, Rajesh and Howard are in the queue for a “Double Cola” drink vending machine; there are no other people in the queue. The first one in the queue (Sheldon) buys a can, drinks it and doubles! The resulting two Sheldons go to the end of the queue. Then the next in the queue (Leonard) buys a can, drinks it and gets to the end of the queue as two Leonards, and so on. This process continues ad infinitum.

For example, Penny drinks the third can of cola and the queue will look like this: Rajesh, Howard, Sheldon, Sheldon, Leonard, Leonard, Penny, Penny.

Write a program that will print the name of a man who will drink the n-th can.

Note that in the very beginning the queue looks like that: Sheldon, Leonard, Penny, Rajesh, Howard. The first person is Sheldon.

Input
The input data consist of a single integer n (1 ≤ n ≤ 109).

It is guaranteed that the pretests check the spelling of all the five names, that is, that they contain all the five possible answers.

Output
Print the single line — the name of the person who drinks the n-th can of cola. The cans are numbered starting from 1. Please note that you should spell the names like this: “Sheldon”, “Leonard”, “Penny”, “Rajesh”, “Howard” (without the quotes). In that order precisely the friends are in the queue initially.

Examples:
input
1
output
Sheldon

input
6
output
Sheldon

input
1802
output
Penny

code 01:

#include<stdio.h>
int main()
{
    int n;
    scanf("%d",&n);
    char *str[]= {"Sheldon","Leonard","Penny","Rajesh","Howard"};
    int round=1;
    while(n>5*round)
    {
        n=n-5*round;
        round*=2;
    }
    printf("%s",str[(n+round-1)/round-1]);
}

code 10:

#include<stdio.h>
#include<math.h>
char*Str[]= {"Sheldon","Leonard","Penny","Rajesh","Howard"};
int main()
{
    float n;
    scanf("%f",&n);
    int k=ceilf(log2f(n/5+1))-1,p=ceilf((float)(n-5*((1<<k)-1))/(1<<k));
    printf("%s\n",Str[p-1]);
}