UVA1363 LA3521 POJ2800 ZOJ2646 Joseph‘s Problem【约瑟夫环+数学】

Joseph’s Problem
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 7906 Accepted: 2107

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 dor := (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 dor := 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 <= 10⁹).

Output

Output the sum requested.

Sample Input

5 3

Sample Output

Source

Northeastern Europe 2005

问题链接：UVA1363 LA3521 POJ2800 ZOJ2646 Joseph’s Problem
问题简述：（略）
问题分析：约瑟夫环问题有关的数学题，参见参考链接，不解释。
程序说明：（略）
参考链接：UVA11351 Last Man Standing【约瑟夫环+数学】
题记：（略）

AC的C++语言程序如下：

/* UVA1363 LA3521 POJ2800 ZOJ2646 Joseph's Problem */#include <iostream>
#include <cstdio>
#include <cmath>using namespace std;typedef long long LL;LL jose(LL n, LL k)
{LL sum = 0, a = sqrt(k), b = k / a;if(n > k) sum += (n - k) * k;for(LL i = a; i >= 2; i--) {LL start = k / i;LL end2 = k / (i - 1);if(start > n) break;if(end2 > n) end2 = n;sum += (k % end2 + k % (start + 1)) * (end2 - start) / 2;}for(LL i = 1; i <= n && i <= b; i++)sum += k % i;return sum;
}int main()
{LL n, k;while(~scanf("%lld%lld", &n, &k))printf("%lld\n", jose(n, k));return 0;
}

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UVA1363 LA3521 POJ2800 ZOJ2646 Joseph‘s Problem【约瑟夫环+数学】

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