zoukankan      html  css  js  c++  java
  • hdu-1517-博弈

    A Multiplication Game

    Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 6917    Accepted Submission(s): 3930


    Problem Description
    Stan and Ollie play the game of multiplication by multiplying an integer p by one of the numbers 2 to 9. Stan always starts with p = 1, does his multiplication, then Ollie multiplies the number, then Stan and so on. Before a game starts, they draw an integer 1 < n < 4294967295 and the winner is who first reaches p >= n.
     
    Input
    Each line of input contains one integer number n.
     
    Output
    For each line of input output one line either 

    Stan wins. 

    or 

    Ollie wins.

    assuming that both of them play perfectly.
     
    Sample Input
    162 17 34012226
     
    Sample Output
    Stan wins. Ollie wins. Stan wins.
     
    Source
     
        我们可以逆推出必胜区间和必败区间,一段一段递推,当边界到达1的时候输出此时的状态就好了,初始化L=n,op=LOSE,
    当上一个区间状态是LOSE的时候,就执行 L=ceil(L/9) op=WIN
    反之执行  L=ceil(L/2) op=LOSE.
     
    只要记住1只要有一个子状态到0,而0的所有子状态都必须是1即可。
     1 #include<bits/stdc++.h>
     2 using namespace std;
     3 int main(){
     4     long long n;
     5     while(cin>>n){
     6         bool op=0;
     7         while(n>1){
     8             if(op==0){
     9                 op=1;
    10                 n=ceil((double)n/9);
    11             }
    12             else{
    13                 op=0;
    14                 n=ceil((double)n/2);
    15             }
    16             
    17         }
    18         op?puts("Stan wins."):puts("Ollie wins.");
    19     }
    20     return 0;
    21 }
  • 相关阅读:
    CF431E Chemistry Experiment
    BZOJ 4173: 数学
    BZOJ 2426: [HAOI2010]工厂选址
    BZOJ 2580: [Usaco2012 Jan]Video Game
    BZOJ 4237: 稻草人
    BZOJ 2434: [Noi2011]阿狸的打字机
    BZOJ 3881: [Coci2015]Divljak
    BZOJ 2754: [SCOI2012]喵星球上的点名
    BZOJ 1009: [HNOI2008]GT考试
    BZOJ 3731: Gty的超级妹子树
  • 原文地址:https://www.cnblogs.com/zzqc/p/9350196.html
Copyright © 2011-2022 走看看