poj 1218 THE DRUNK JAILER【水题】

THE DRUNK JAILER

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 25124		Accepted: 15767

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

题意：第一个人把所有的牢房全部打开，第二个人把是2的倍数的牢房全部锁上，第三个人对是3的倍数的牢房进行操作，如果是锁上的就把它打开，如果是打开的就把它锁上，依次第4个人...直到第n个人完成操作，问最后有多少间牢房是打开的

#include<stdio.h>
#include<string.h>
int a[110];
int main()
{
int t,n,i,j;
scanf("%d",&t);
while(t--)
{
int sum=0;
scanf("%d",&n);
memset(a,0,sizeof(a));
for(j=1;j<=n;j++)
{
for(i=1;i<=n;i++)
{
if(i%j==0)
{
if(a[i]==0)
a[i]=1;
else if(a[i]==1)
a[i]=0;
}
}
}
for(i=1;i<=n;i++)
if(a[i]==1)
sum++;
printf("%d\n",sum);
}
return 0;
}