And Then There Was One （约瑟夫环变形）

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 khops 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
解题思路：有n个石头围成一圈，第一次移走第m个石头，然后从第m+1个石头从1开始数，以后每次数到k就移走一个石头，第k+1个石头又从1开始数，依此规律重复下去，求最后一个移走的石头编号。做法：假设编号为0~n-1的n个石头围成一圈，从0开始每k个石头移走一个，最后留下的编号记为f[n]。因为第一次移走第k-1个石头是从第0个石头开始数的，而第一次移走第m-1个石头是从第m-k个石头开始数的，所以只需将原来0~n-1重新编号后可以得到：最终剩下一个石头的编号为(f[n]+m-k)%n，因为f[n]加上偏移量后可能为负或者超过n，所以最后应该加上n再取模n，即((f[n]+m-k)%n+n)%n，这就是第一次移走第m个石头的最终结果。
AC代码：

#include<iostream>
#include<cstdio>
using namespace std;
int n,k,m,s;
int main(){
    while(~scanf("%d%d%d",&n,&k,&m)&&(n+k+m)){
        s=0;//只有一个石头，移走的编号为0
        for(int i=2;i<=n;++i)s=(s+k)%i;
        s=((s+m-k+n)%n+n)%n;
        cout<<(s+1)<<endl;//因为计算是从0开始的，所以最终的编号要加1
    }
    return 0;
}