UVA 10110 (13.07.26)

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题意:一个走廊上有n个灯, 一开始是全部关闭状态一个神经病, 走n次走廊, 假如n是6, 那么6的因子数有1, 2, 3, 6.那么走四次, 第一次每隔1盏灯就转化对应灯状态, 第二次每隔2盏灯转化状态, 以此类推~那么很容易看出来, n盏灯, 除非是平方数, 不然其因子数是成对出现, 如6中的 1和6, 2和3而平方数多了一个开方的因子数, 其因子数就变成了奇数个, 从而状态改变就是由开始的关变为了开注意:用int, 理论上可行, 实际上我WA了, 改用long long过了做法:不说了...AC代码:
#include<stdio.h>#include<math.h>int main() {long long n, k;while(scanf("%lld", &n) != EOF) {if(n == 0)break;k = sqrt(n);if(k * k == n)printf("yes\n");elseprintf("no\n");}return 0;}