zoukankan      html  css  js  c++  java
  • And Then There Was One

    http://poj.org/problem?id=3517

    And Then There Was One
    Time Limit: 5000MS   Memory Limit: 65536K
    Total Submissions: 4805   Accepted: 2546

    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


     1 #include"string.h"
     2 int f[10001];
     3 int main()
     4 {
     5     int n,k,m,i,j;
     6     while(scanf("%d%d%d",&n,&k,&m)==3&&n)
     7     {
     8         f[1]=0;
     9         for(i=2;i<n;i++)   //重新编号,即编号映射 
    10             f[i]=(f[i-1]+k)%i;
    11         f[n]=(f[n-1]+m)%n;
    12         printf("%d
    ",f[n]+1); 
    13     }
    14     return 0;
    15 }
  • 相关阅读:
    C++设计模式之享元模式
    C++设计模式之中介者模式
    C++设计模式之职责链模式
    C++设计模式之命令模式
    C++设计模式之桥接模式
    C++设计模式之单例模式
    C++设计模式之组合模式
    C++设计模式之备忘录模式
    C++设计模式之适配器模式
    操作系统(4)实验0——准备知识、基本内联汇编、扩展内联汇编
  • 原文地址:https://www.cnblogs.com/767355675hutaishi/p/4009165.html
Copyright © 2011-2022 走看看