boj 447

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: 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


Source
tokyo 2007

 

约瑟夫环问题，n个人编号1~n围成一圈报数，报到m的人出列，循环，找最后剩下的人编号。

初始序列为：1,2,3,4,5...m-1,m,m+1...n

去掉一个人后:1,2,3,4,5...m-1,m+1...n

m+1,m+2...n...1,2,3,...m-1

做如下变换：

m+1->1

m+2->2

....

....

m-1->n-1

则序列为：1,2,3,4，...n-1

变成了n-1个人报数的问题，递归求解，设n-1个人的解为x，易得在原始序列中的解为（x+m）%n(注意模为0的情况，即解为n);

此题要注意的就是第一次是报m，后面全部报k

代码：

#include<iostream>
using namespace std;
int main()
{
	int n,k,m;
	scanf("%d %d %d",&n,&k,&m);
	while(!(n==0&&m==0&&k==0))
	{
		int tmp = 1;
		for(int i =2;i<n;i++)
		{
			tmp = (tmp+k)%i;
			if(tmp==0)
				tmp = i;
		}
		tmp = (tmp+m)%n;
		if(tmp==0)
			tmp = n;
		printf("%d\n",tmp);
		scanf("%d %d %d",&n,&k,&m);
	}
}