And Then There Was One

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.

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.

2n10000, 1k10000, 1mn

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

题意:将石头排序,将石头m拿掉,从石头m开始,每经过K的石头,拿掉第k个...求最后剩下哪一个...

思路:约瑟夫环问题变型..

     原型是从第一个石头开始的...

     1.将石头编号改为从0到n-1,mod n,再加1,可得到结果..

     2.递推公式 f[i]=(f[i-1]+k)%i,f[1]=0; f[i]表示第i个人时胜利者的编号...

       最后一步应该分开来算..因为石头不是从1 开始算起的,而是第m个石头..

        f[n]=(f[n-1]+m)%n;

CODE:

#include<stdio.h>#include<iostream>using namespace std;int f[10005];int main(){    int n,k,m;    while(~scanf("%d%d%d",&n,&k,&m),n,k,m)    {       f[1]=0;       for(int i=2;i<n;i++)            f[i]=(f[i-1]+k)%i;       f[n]=(f[n-1]+m)%n+1;       printf("%d\n",f[n]);    }    return 0;}