UVA 10110 Light, more light

Light,more light

The Problem

There is man named "mabu" for switching on-off light in our University.He switches on-off the lights in a corridor. Every bulb has its own toggle switch. That is, if it is pressed then the bulb turns on. Another press will turn it off. To save power consumption(or may be he is mad or something else) he does a peculiar thing. If ina corridor there is `n' bulbs, he walks along the corridor back and forth`n' times and in i'th walk he toggles only the switches whose serial is divisable by i. He does not press any switch when coming back to his initial position.A i'th walk is defined as going down the corridor (while doing the peculiar thing) and coming back again.

Now you have to determine what is the final condition of the last bulb. Is it on or off?
 

The Input

The input will be an integer indicating the n'th bulb in a corridor. Which is less thenor equals 2^32-1. A zero indicates the end of input. You should not processthis input.

The Output

Output "yes" if the light is on otherwise "no" , in a single line.

SampleInput

3624181910

SampleOutput

noyesno

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Sadi Khan
Suman Mahbub
01-04-2001

题目意思就是有一个走廊，走廊上有很多灯。。有一个人在那开关。。

如过有ｎ盏灯，他就走ｎ躺，回来的时候不动开关。。

走第一趟，就动可以被一整除的开关，

走第二趟就动能被二整除的开关，第２，４，６，８．。。。盏。。

问ｎ趟走完，最后一盏灯是开是关。。。

如果从１开始遍历，看看能被整出的数有多少个，偶数个就是关，奇数个就是开的话，，是一定会超时的。。。

所以找到了一个规律，，如果灯的数量是可以被开方的，，像１，４，９，１６最后就是开的，ｙｅｓ，否则就是ｎｏ。。

ＡＣ代码：

#include<stdio.h>#include<cmath>int main () {double num;while (scanf("%lf",&num)) {if (num == 0)break;if ( (int)sqrt(num) == sqrt(num)  )printf("yes\n");else printf("no\n");}return 0;}