HDU1337(hash)

最新推荐文章于 2017-03-12 15:37:14 发布

晓风残月xj

最新推荐文章于 2017-03-12 15:37:14 发布

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分类专栏：数据结构-hash 文章标签： hash

本文链接：https://blog.youkuaiyun.com/xiaofengcanyuexj/article/details/8882265

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本文探讨了一种独特的监狱逃狱游戏，通过特定的监狱管理规则，计算囚犯逃脱的可能性。游戏涉及监狱长在不同轮次中解锁和锁定囚犯的细胞，最终导致某些囚犯得以逃脱。通过数学逻辑分析，我们能够计算出在给定的细胞数量下，有多少囚犯能够成功逃脱。

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The Drunk Jailer

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 885 Accepted Submission(s): 735

Problem Description

A certain prison contains a long hall of n cells, each right next to each other. Each cell has a prisoner in it, and each cell is locked.
One night, the jailer gets bored and decides to play a game. For round 1 of the game, he takes a drink of whiskey, and then runs down the hall unlocking each cell. For round 2, he takes a drink of whiskey, and then runs down the hall locking every other cell (cells 2, 4, 6, …). For round 3, he takes a drink of whiskey, and then runs down the hall. He visits every third cell (cells 3, 6, 9, …). If the cell is locked, he unlocks it; if it is unlocked, he locks it. He repeats this for n rounds, takes a final drink, and passes out.

Some number of prisoners, possibly zero, realizes that their cells are unlocked and the jailer is incapacitated. They immediately escape.

Given the number of cells, determine how many prisoners escape jail.

Input

The first line of input contains a single positive integer. This is the number of lines that follow. Each of the following lines contains a single integer between 5 and 100, inclusive, which is the number of cells n.

Output

For each line, you must print out the number of prisoners that escape when the prison has n cells.

Sample Input

2
5
100

Sample Output

2
10

hash可以有效减少循环次数，牺牲空间换取时间

#include<iostream>
using namespace std;
int hash[110];

int main()
{
 int i,j,t,n,num;
 cin>>t;
 while(t--)
 {
  num=0;
  cin>>n;
  memset(hash,0,sizeof(hash));
  for(i=1;i<=n;i++)
  {
   for(j=i;j<=n;j=j+i)
   {
    if(hash[j]==0)
     hash[j]=1;
    else 
     hash[j]=0;
   }
  }

  for(i=1;i<=n;i++)
  {
   if(hash[i]==1)
    num++;
  }

  cout<<num<<endl;
 }
 return 0;
}