zoukankan      html  css  js  c++  java
  • Virtual Judge —— Nim TopCoder

    Problem Statement 问题陈述

    Alice and Bob are going to play a famous game called Nim.
    爱丽丝和鲍勃将要玩一个著名的游戏叫尼姆。
    In the game Nim, first they set up stones in K piles containing a1,…,aK stones respectively.
    在尼姆游戏中,首先在K堆中分别设置了包含a1,…,aK的石头。
    Then they alternatively take turns (Alice moves first).
    然后他们轮流移动(爱丽丝先移动)。
    On a player’s turn the player chooses a pile and takes some (at least one) stones from that pile.
    轮到玩家时,玩家选择一堆石头,从那堆石头中取出一些(至少一块)。
    If there are no piles left which contain any stones, the player loses.
    如果没有剩余的包含任何石头的桩,玩家将失败。

    Since they like prime numbers very much, they decided to make each ai a prime number less than or equal to L.
    因为他们非常喜欢素数,所以他们决定使每个ai都成为小于或等于L的素数。
    Given K and L return the number of such initial setups which allows Bob to win, assuming they play optimally, modulo 1,000,000,007.
    给定k和l返回允许Bob获胜的初始设置的数量,假设它们发挥的最佳,模100000000007。

    Definition 定义

    Class: Nim
    类别:尼姆
    Method: count
    方法:计数
    Parameters: int, int
    参数:int,int
    Returns: int
    返回:int
    Method signature: int count(int K, int L)
    方法签名:int count(int k,int l)
    (be sure your method is public)
    (确保您的方法是公开的)

    Notes 笔记

    • Two setups are considered different if at least one ai is different between them (for example, (a1,a2,a3)=(2,5,7) and (2,7,5) are considered different).
    • 如果两个设置之间至少有一个AI不同(例如,(A1、A2、A3)=(2、5、7)和(2、7、5)视为不同),则认为两个设置不同。

    Constraints 约束条件

    • K will be between 1 and 1000000000(=109), inclusive.
    • K包含在1和1000000000之间。
    • L will be between 2 and 50000, inclusive.
    • L包含在2和50000之间。

    Examples 样例

    3
    7
    Returns: 6
    Prime numbers <= 7 are 2, 3, 5 and 7. Bob can win if the initial setup is (2,5,7) or its permutation. So return 3! = 6.

    4
    13
    Returns: 120
    Bob can win if the initial setup is (p,p,p,p) for some prime p<=13, (p,p,q,q) or its permutation for p<q<=13, or (3,5,11,13) or its permutation. So return 6+(6C2*6)+4!=6+90+24=120.

    10
    100
    Returns: 294844622

    123456789
    12345
    Returns: 235511047
    This problem statement is the exclusive and proprietary property of TopCoder, Inc. Any unauthorized use or reproduction of this information without the prior written consent of TopCoder, Inc. is strictly prohibited. ©2010, TopCoder, Inc. All rights reserved.

  • 相关阅读:
    取最小未用的账号
    Oracle 同义词
    RMAN备份报 RMAN-06059 错误
    Oracle11g memory_target
    基础记录
    uboot启动流程
    android有关生命周期探讨
    can协议
    java虚拟机
    内存管理机制
  • 原文地址:https://www.cnblogs.com/AlexKing007/p/12338379.html
Copyright © 2011-2022 走看看