Hackerearth.com编程问题解题思路系列：Roy's Chocolates

问题描述：

设从1至n编号的n人围成一圈，给出序号M[i]的m人是被标记的，从序号为b的人开始，每过s人答一声“到”。要求找出这样的b和s，使得被标记的m人全部都答过一声“到”，而没有被标记过的人不答“到”。n在10^9量级，m在10^6量级。

问题分析：

最原始的方法是枚举s和b，时间复杂度为O(n^2*m)，由于n，m的量级比较大，所以不能直接硬举，要先寻找一些规律，再配合搜索进行求解。

首先将M[i]在坐标轴上画下来：将1到n向右在坐标轴上扩展，令M1,M2...Mm分别为第1，2...，m个答到的人，按每隔s的顺序在坐标轴上标出，如下图：



注：图中相邻两个M之间距离为s

经观察得出：



将上面所有式相加，并在两边各加上M1，得到：



式中m与各个Mi之和是已知的，M1即是出发点b，是枚举的对象，还差一个s没有确定，如果枚举s，可以先确定出s的范围。

这时将公式改写成如下形式：



而t是可以确定的：



[x]为取整函数(不超过x的最大整数)

于是可得：



整理得：



这时s的范围可以定得比较小，枚举s基本上花费常数时间，算法整体的时间复杂度是O(n)

Python代码：

n,m=[int(i) for i in raw_input().split()]
M=[int(i) for i in raw_input().split()]
def calc():
global n,m
global M
if m==1:
print "%d %d"%(M[0],n)
return
sumM=sum(M)
for i in range(m):
left=int((sumM-m*M[i]-n)*2/(m-2)/(m-1))
if left<=0:
left=1
right=int((sumM-m*M[i])*2/(m-2)/(m-1))+1
#print left,right
for l in range(left,right+1):
if int((m*M[i]+m*(m-1)*l/2))%n==sumM%n and (M[i]+l) in M and (M[i]+2*l) in M:
print "%d %d"%(M[i],l)
return
print "impossible"
calc()


原题：https://www.hackerearth.com/problem/approximate/roys-chocolates/

Roy has brought one chocolate for each of his m friends to class today. After knowing that, all of his n classmates demanded chocolates from him. But Roy wanted to give
chocolates only to his friends.

So, he told them to stand in a circle around him, where the 1st student stands next to the nth one. Then Roy will follow a strategy to distribute
the chocolates. He will start by giving chocolate to some person b, and will continue to give chocolate to every sth person. After someone receives chocolate,
s/he does not leave the circle.

Roy noticed the positions of his friends, and now he wants to choose b and s in such a way that, only and all of his friends get those chocolates.

Input Format

The first line of input contains 2 space separated integers, n and m, number of Roy's classmates and number of Roy's friends.
The next line contains m space separated integers, the positions of Roy's friends.

Output Format

Output suitable b and s to execute Roy's plan. If no suitable answer exists, output "impossible" ( without the quotes ).

If there are multiple solutions, print any of them.

Constraints

1 ≤ n ≤ 10 9 
1 ≤ m ≤ 10 6 
m ≤ n 

positions of Roy's friends will be given in increasing order.