And Then There Was One （约瑟夫环变形）

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.
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

题目大概：

给出n个人成环，报数从m开始，每次报数间隔k个人。

思路：

正常约瑟夫环是从1号开始，如果从m号，那么就是报数延迟到了m，那最后报数的人也会相应延迟m。

代码：

#include <bits/stdc++.h>
using namespace std;
const int maxn=11000;
int dp[maxn];

int main()
{
    int n,k,m;
    while(~scanf("%d%d%d",&n,&k,&m))
    {
        if((n+k+m)==0)break;
        dp[1]=0;
        for(int i=2;i<n;i++)
        {
            dp[i]=(dp[i-1]+k)%i;
        }
        printf("%d\n",(dp[n-1]+m)%n+1);
    }
    return 0;
}