首页 > ACM题库 > HDU-杭电 > HDU 1337 The Drunk Jailer-模拟-[解题报告] C++
2013
12-09

HDU 1337 The Drunk Jailer-模拟-[解题报告] C++

The Drunk Jailer

问题描述 :

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.

输入:

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.

输出:

For each line, you must print out the number of prisoners that escape when the prison has n cells.

样例输入:

2
5
100 

样例输出:

2
10

地址:http://acm.hdu.edu.cn/showproblem.php?pid=1337

题意:狱警玩游戏,第一次转换1,2,……,n的锁的开关状态,第二次转换2,4,……,第三次3,6,……依次类推到第n次。求最后有多少锁是开着的。

mark:模拟就ok了。

代码:

#include <stdio.h>
#include <string.h>

int main()
{
    int t,n,a[110];
    int i,j,sum;
    scanf("%d", &t);
    while(t-- && scanf("%d", &n))
    {
        memset(a, 0, sizeof(a));
        for(i = 2; i <= n; i++)
            for(j = 1; j*i <= n; j++)
                a[j*i]++;
        for(sum = 0, i = 1; i <= n; i++)
            if(!(a[i]%2)) sum++;
        printf("%d\n", sum);
    }
    return 0;
}

解题报告转自:http://www.cnblogs.com/andre0506/archive/2012/07/03/2574059.html


  1. 算法是程序的灵魂,算法分简单和复杂,如果不搞大数据类,程序员了解一下简单点的算法也是可以的,但是会算法的一定要会编程才行,程序员不一定要会算法,利于自己项目需要的可以简单了解。

  2. #!/usr/bin/env python
    def cou(n):
    arr =
    i = 1
    while(i<n):
    arr.append(arr[i-1]+selfcount(i))
    i+=1
    return arr[n-1]

    def selfcount(n):
    count = 0
    while(n):
    if n%10 == 1:
    count += 1
    n /= 10
    return count

  3. [email protected]