And Then There Was One（DP）

 

And Then There Was One

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 4509		Accepted: 2384

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

 

    题意：

    给出 N（2 ~ 10000），K（1 ~ 10000），M（1 ~ N）。代表有 1 ~ N 个数，每隔 K 个数就删除一个数，从 M 这个数开始数（M 这个数第一删除）。输出最后一个输出的数是什么。属于多组输入。

 

    思路：

    递推。DP。

    假设K = 5（不考虑第一个数被删除）：

    N = 2 时，即 1， 2  => 从1 开始数，1,2,1,2,1，删除1，故最后删除的是 2，表示当 N = 2 时，最后将会删除第二个数；

    N = 3 时，即 1，2，3 => 从 1 开始数，1,2,3,1,2，删除2，还剩两个数 N = 2，由上面的情况可得知，最后将会删除接下来数的顺序的第二个数，故最后会删去第一个数；

    由此递推下去，可得知推知结果。

    与此考虑不同只是第一删除的会是 M 这个元素，故不考虑该元素，从下个元素开始即可。

 

    AC：

#include<stdio.h>
int fin[10005];
int main()
{
    int n,k,m;
    while(~scanf("%d%d%d",&n,&k,&m) && (n + k + m))
    {
        int num;
        fin[2] = (k % 2) ? 2 : 1;
        for(int i = 3;i <= n - 1;i++)
        {
            int ans;
            if(i < k) ans = k % i;
            else      ans = k;
            fin[i] = (ans + fin[i - 1]) % i;
            if(!fin[i]) fin[i] = i;
        }
        if(n > 2)
        {
            num = (fin[n - 1] + m) % n;
            if(!num)  num = n;
        }
        else
        {
            if(m == 1) num = 2;
            if(m == 2) num = 1;
        }
        printf("%d\n",num);
    }
    return 0;
}