ZOJ Problem Set–1350 The Drunk Jailer

Time Limit: 2 Seconds      Memory Limit: 65536 KB

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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

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Source: Greater New York 2002

这题是考循环控制的，初学程序设计的人可以做一下，增强对循环的理解。

#include<iostream>
#include<bitset>
using namespace std;
int main()
{
  int lines,n;cin>>lines;
  while(lines-- && cin>>n)
  {
    bool *pCells = new bool[n];
    bool locked = false;
    for(int i = 0; i < n; i++)//初始化所有监狱，在第一轮中所有监狱被打开
    {
      *(pCells + i) = locked;
    }
    for(int j = 2; j <= n;j++)
    {
      for(int k = j; k <= n;k += j)
      {
        *(pCells + k - 1) = !*(pCells + k - 1);
      }
    }
    int unlockedCell = 0;
    for(int z = 0; z < n; z++)
    {
      if(!*(pCells + z))
        unlockedCell++;
    }
    cout<<unlockedCell<<endl;
    delete []pCells;
  }
  return 0;
}