UVa 10110 Light, more light:数论

10110 - Light, more light

Time limit: 3.000 seconds

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=100&page=show_problem&problem=1051

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

注意一开始所有灯泡都是关着的。

简化：我们只需要计算满足i|n的i的个数(即n的因子数)的奇偶性即可。

注意到当n不是完全平方数的时候，n=a*b中a一定不等于b，此时n有偶数个因子。

反之当n为完全平方数时，n一定有奇数个因子。

完整代码：

/*0.039s*/

#include<cstdio>
#include<cmath>  

int main()
{
    unsigned int n;///注意数据范围
    int m;
    while (scanf("%d", &n), n)
    {
        m = (int)sqrt(n);
        puts(m * m == n ? "yes" : "no");
    }
    return 0;
}

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