POJ 2800 Joseph's Problem

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最新推荐文章于 2018-03-21 19:42:56 发布

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本文探讨了一个基于Joseph问题的变种问题，通过分析并提供了一种有效的算法来解决该问题。核心内容包括如何通过数学方法简化求和过程，并给出了解决方案的具体实现代码。

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给n 和 k 求: ∑_1<=i<=n(k mod i).

p = k/i

k mod i = k - p * i

k mod ( i + 1 ) = k - p * ( i + 1 ) = k mod i - p

k mod ( i + 2 ) = k - p * ( i + 2 ) = k mod i - 2 * p

对于连续的 i ,很多p都是一样的 . 相差的部分是一个等差数列 ,

i 的范围是从 i 到 min(k/p,n) 如果 p == 0 则一直延续到最后

Joseph's Problem

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7282		Accepted: 1891

Description

Joseph likes taking part in programming contests. His favorite problem is, of course, Joseph's problem.

It is stated as follows.

There are n persons numbered from 0 to n - 1 standing in a circle. The person numberk, counting from the person number 0, is executed. After that the person number k of the remaining persons is executed, counting from the person after the last executed one. The process continues until only one person is left. This person is a survivor. The problem is, given n and k detect the survivor's number in the original circle.

Of course, all of you know the way to solve this problem. The solution is very short, all you need is one cycle:

	r := 0;

	for i from 1 to n do

		r := (r + k) mod i;

	return r;

Here "x mod y" is the remainder of the division of x by y, But Joseph is not very smart. He learned the algorithm, but did not learn the reasoning behind it. Thus he has forgotten the details of the algorithm and remembers the solution just approximately.

He told his friend Andrew about the problem, but claimed that the solution can be found using the following algorithm:

	r := 0;

	for i from 1 to n do

		r := r + (k mod i);

	return r;

Of course, Andrew pointed out that Joseph was wrong. But calculating the function Joseph described is also very interesting.

Given n and k, find ∑ _1<=i<=n(k mod i).

Input

The input file contains n and k (1<= n, k <= 10 ⁹).

Output

Output the sum requested.

Sample Input

5 3

Sample Output

Source

Northeastern Europe 2005

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/* ***********************************************
Author        :CKboss
Created Time  :2015年02月02日 星期一 10时53分55秒
File Name     :POJ2800.cpp
************************************************ */

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>

using namespace std;

typedef long long int LL;

LL n,k;

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);

	while(cin>>n>>k)
	{
		LL ret=0;
		for(int i=1;i<=n;i++)
		{
			LL p = k/i;
			if(p==0)
			{
				ret += (n-i+1)*k;
				break;
			}
			else
			{
				LL q = k/p;
				if(q>n) q = n;
				LL md = k%i;
				LL num = q-i+1;
				LL temp = md*num-(num-1)*num/2*p;
				ret += temp;
				i=q;
			}
		}
		cout<<ret<<endl;
	}
    
    return 0;
}