3517 And Then There Was One 约瑟夫问题

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最新推荐文章于 2024-08-26 00:22:36 发布

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分类专栏： acm_数学问题文章标签： dataset numbers output input each less

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本文介绍了一个有趣的算法问题：从一圈编号的石头中，遵循特定规则移除石头，直至剩下最后一块。通过示例详细解释了游戏规则，并提供了一段C++代码实现。

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And Then There Was One

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 3136		Accepted: 1562

Description

Let’s play a stone removing game.

Initially, n stones are arranged on a circle and numbered 1, …, n clockwise (Figure 1). You are also given two numbers k and m. From this state, remove stones one by one following the rules explained below, until only one remains. In step 1, remove stone m. In step 2, locate the k-th next stone clockwise from m and remove it. In subsequent steps, start from the slot of the stone removed in the last step, make k hops clockwise on the remaining stones and remove the one you reach. In other words, skip (k − 1) remaining stones clockwise and remove the next one. Repeat this until only one stone is left and answer its number. For example, the answer for the case n = 8, k = 5, m = 3 is 1, as shown in Figure 1.

Initial state	Step 1	Step 2	Step 3	Step 4
Step 5	Step 6	Step 7	Final state

Figure 1: An example game

Initial state: Eight stones are arranged on a circle.

Step 1: Stone 3 is removed since m = 3.

Step 2: You start from the slot that was occupied by stone 3. You skip four stones 4, 5, 6 and 7 (since k = 5), and remove the next one, which is 8.

Step 3: You skip stones 1, 2, 4 and 5, and thus remove 6. Note that you only count stones that are still on the circle and ignore those already removed. Stone 3 is ignored in this case.

Steps 4–7: You continue until only one stone is left. Notice that in later steps when only a few stones remain, the same stone may be skipped multiple times. For example, stones 1 and 4 are skipped twice in step 7.

Final State: Finally, only one stone, 1, is on the circle. This is the final state, so the answer is 1.

Input

The input consists of multiple datasets each of which is formatted as follows.

n k m

The last dataset is followed by a line containing three zeros. Numbers in a line are separated by a single space. A dataset satisfies the following conditions.

2 ≤ n ≤ 10000, 1 ≤ k ≤ 10000, 1 ≤ m ≤ n

The number of datasets is less than 100.

Output

For each dataset, output a line containing the stone number left in the final state. No extra characters such as spaces should appear in the output.

Sample Input

8 5 3
100 9999 98
10000 10000 10000
0 0 0

Sample Output

1
93
2019

#include<iostream>
#include<cstdio>
using namespace std;
int main()
{
    int n,k,m;//第一次move第m个，剩下的move后数第k个
    while(scanf("%d%d%d",&n,&k,&m)==3&&n)
    {
        int s=0;
        for(int i=2;i<=n;i++)
        {
            if(i==n) s=(s+m)%i;
            else s=(s+k)%i;
        }
        printf("%d/n",s+1);
    }
    return 0;
}