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原創(chuàng)：Light, more light - PC110701

作者： MilkCu

題目描述

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 in a 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 then or equals 2^32-1. A zero indicates the end of input. You should not process this input.

The Output

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

Sample Input

    3
6241
8191
0
  

Sample Output

    no
yes
no

  

---

Sadi Khan ?
Suman Mahbub ?
01-04-2001

解題思路

任一整數(shù)，它可能為平方數(shù)或非平方數(shù)。
若為非平方數(shù)，則它的因子總是成對出現(xiàn)的，即因子個數(shù)為偶數(shù)；
若為平方數(shù)，則它的因子個數(shù)為奇數(shù)。
也就是說：
如果n為非平方數(shù)，則燈的狀態(tài)為關(guān)；
如果n為平方數(shù)，則燈的狀態(tài)為開。

剛開始評測一直說是錯誤的答案，可能是因為數(shù)據(jù)類型和頭文件的問題。

代碼實現(xiàn)

    #include <iostream>
#include <cmath>
#include <algorithm>
using namespace std;
int main(void) {
	while(1) {
		long long n;
		cin >> n;
		if(n == 0) {
			break;
		}
		long long tmp = sqrt(n);
		if(n == tmp * tmp) {
			cout << "yes" << endl; 
		} else {
			cout << "no" << endl;
		}
	}
	return 0;
}
  

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本文地址： http://blog.csdn.net/milkcu/article/details/23543557

Light, more light - PC110701