POJ 1218 THE DRUNK JAILER

THE DRUNK JAILER
Time Limit: 1000MS   Memory Limit: 10000K
Total Submissions: 21957   Accepted: 13879

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

Source

和UVa 10110基本相同,只不过这个是求总个数,而Uva那道只要判断最后一个

解题思路参见UVa 10110(http://www.cnblogs.com/lzj-0218/p/3531714.html

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cmath>
 4 
 5 using namespace std;
 6 
 7 int main()
 8 {
 9     int kase;
10 
11     scanf("%d",&kase);
12 
13     while(kase--)
14     {
15         int n;
16         scanf("%d",&n);
17         int tmp=sqrt(n+0.5);
18         printf("%d
",tmp);
19     }
20 
21     return 0;
22 }
[C++]
原文地址:https://www.cnblogs.com/lzj-0218/p/3537105.html