## 開燈問題

`7 3`

`1 5 6 7`
`<pre name="code" class="cpp">#include<stdio.h>#include<string.h>int a[10000]; //數組要開大點 int main(){int n,m;while(~scanf("%d %d",&n,&m))//有n個數，進行幾輪，第一輪都開開門，第二輪關上（打開）是二的倍數，以此類推 {memset(a,0,sizeof(a));//數組清空，開門為0，所以，第一輪都開門。 int i,k,j;for(i=2;i<=m;i++){k=i;for(j=1;k<=n;j++){k=i*j;//找出第m輪m的倍數，並檢查是否在n的范圍內 if(a[k]==0)a[k]=-1;//關門為-1 elsea[k]=0;}}printf("1");for(i=2;i<=n;i++){if(a[i]==0)printf(" %d",i);}printf("\n");}return 0;} `
THE DRUNK JAILER
 Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 25125 Accepted: 15768

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

`25100`

Sample Output

`210`
`<span style="color:#464646;">#include<stdio.h>#include<string.h>int main(){int n,t,a[1000],i,j,k;scanf("%d",&t);while(t--){scanf("%d",&n);memset(a,0,sizeof(a));for(i=2;i<=n;i++){k=i;for(j=1;k<=n;j++){k=i*j;if(a[k]==0)a[k]=-1;elsea[k]=0;}}int j=0;for(i=1;i<=n;i++){if(a[i]==0)j++;}printf("%d\n",j);}return 0;}                                                                                                                                                                                                                                                                                                      `
`#include<stdio.h>#include<math.h>int main(){int T,n,m;scanf("%d",&T);while(T--){scanf("%d",&n);m=sqrt(n);printf("%d\n",m);}return 0;}`