UVa 10110 Light, more light

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開始所有的燈是滅的，不過我們只關心最後一個燈。

在第i次走動時，只有編號為i的倍數的燈的狀態才會改變。

也就是說n有偶數個約數的時候，最後一個燈的狀態不會改變，也就是滅的。

n有奇數個約數的時候也就是n為完全平方數的時候，最後一個燈會是亮的。

 

最後抽象出來，就是判斷輸入的數是否為完全平方數。

 

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

3624181910

Sample Output

noyesno

Sadi Khan 
Suman Mahbub 
01-04-2001

 

AC代碼：

 1 //#define LOCAL 2 #include <iostream> 3 #include <cstdio> 4 #include <cstring> 5 #include <cmath> 6 using namespace std; 7  8 int main(void) 9 {10     #ifdef LOCAL11         freopen("10110in.txt", "r", stdin);12     #endif13 14     unsigned int N;15     while(scanf("%d", &N) == 1 && N)16     {17         int n = (int)sqrt(N);18         if(n * n == N)19             cout << "yes" << endl;20         else21             cout << "no" << endl;22     }23     return 0;24 }

代碼君