POJ1218THE DRUNK JAILER 快速和一般方法两种解法

THE DRUNK JAILER

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 22155		Accepted: 14012

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

由于最大才为100，所以就算是用枚举一个个的去改变数值最后也能通过，但是如果好好思考下其实很简单，就是一个数学规律，或许大神和菜鸟的区别就在于此，虽然题目简单但是值得我们去思考和延伸..........

-----------------AC代码----------------------------------

菜鸟方法：

#include<stdio.h>
#define MAX 101
int main()
{
	int a[MAX],x,i,j,k,n,m,y;
	scanf("%d",&m);
	for(y=0;y<m;y++)
	{
	scanf("%d",&n);
	for(i=1;i<=n;i++)
	{
		a[i]=1;
	}
	k=1;
	while(k<=n)
	{
		for(j=k;j<=n;j+=k)
		{
			 a[j]=-a[j];
		}
		k++;
	}
	x=0;
	for(i=1;i<=n;i++)
	  if(a[i]==-1) x++;
	printf("%d\n",x);
    }
	return 0;
}

大神方法：

#include<stdio.h>
#include<stdlib.h>
#include<math.h>


int main()
{
    int n,x,i;
    scanf("%d",&n);
    for(i=0;i<n;i++)
    {
       scanf("%d",&x);
       printf("%d\n",(int)sqrt((double)x));
       
    }
    system("pause");
    return 0;
}