CodeForces - 82A

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.

Example
Input
1
Output
Sheldon
Input
6
Output
Sheldon
Input
1802
Output
Penny

题意描述：

五个人排队去喝汽水，依次是：Sheldon, Leonard, Penny, Rajesh and Howard，每个人喝完之后分裂为两人，这两人从队首回到队尾。
第一次Sheldon喝完后： Leonard , Penny , Rajesh , Howard , Sheldon , Sheldon .
第二次Leonard 喝完后：Penny , Rajesh , Howard , Sheldon , Sheldon ,Leonard,Leonard.
…
…
相当于一个等比数列，有5组，公比为2。

#include<bits/stdc++.h>
//#include<iostream>
//#include<cstdio>
//#include<cmath>
//#include<iomanip>
using namespace std;
//codeforces 82A.Double Cola
/****等比****/
int main(){
int n,n1=0,n2=0;
int temp;
scanf("%d",&n);
if(n>5){     
    n2=n;
    n=n/5+1;   //前n项和公式 a1(1-q^n) / (1-q)=Sn;(2^n-1)/(2-1)*5=Sn;2^n=Sn/5+1
    for(int i=0;i<100;i++){
        if(pow(2,i)>n){  //等比数列前n项和大于输入数据，使用上一组的前n项和
                n1=i-1; //上一组的n
                break;
        }
    }
    temp=n2-(5*(pow(2,n1)-1));  //input-上一组的前n项和
    switch(temp/(int)(pow(2,n1))){   //用temp/分裂的个数 (pow(2,n1))取整
    case 0:
        cout<<"Sheldon"<<endl;
        break;
    case 1:
        cout<<"Leonard"<<endl;
        break;
    case 2:
        cout<<"Penny"<<endl;
        break;
    case 3:
        cout<<"Rajesh"<<endl;
        break;
    case 4:
        cout<<"Howard"<<endl;
        break;
    }
}
else{
    switch(n){
     case 1:
        cout<<"Sheldon"<<endl;
        break;
    case 2:
        cout<<"Leonard"<<endl;
        break;
    case 3:
        cout<<"Penny"<<endl;
        break;
    case 4:
        cout<<"Rajesh"<<endl;
        break;
    case 5:
        cout<<"Howard"<<endl;
        break;
    }
}


return 0;
}

后面在博客上发现了其他的骚操作：

#include <string>
#include <iostream>
using namespace std;

int main(){
    string strs[] = {"Sheldon", "Leonard", "Penny", "Rajesh", "Howard"};
    int n = 0;
    cin>>n;

    int i = 1;
    while (n > i*5)//100 - 5 - 10 - 20 - 40 ...
    {
        n -= i*5;
        i <<= 1;
    }
    int a = n / i;
    if (n % i) a++;
    cout<<strs[a-1];
    return 0;
}