HDU--2053

Switch Game

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 12748 Accepted Submission(s): 7753



[align=left]Problem Description[/align]
There are many lamps in a line. All of them are off at first. A series of operations are carried out on these lamps. On the i-th operation, the lamps whose numbers are the multiple of i change the condition ( on to off and off to
on ).

[align=left]Input[/align]
Each test case contains only a number n ( 0< n<= 10^5) in a line.

[align=left]Output[/align]
Output the condition of the n-th lamp after infinity operations ( 0 - off, 1 - on ).

[align=left]Sample Input[/align]

1
5

[align=left]Sample Output[/align]

1
0

Hinthint

Consider the second test case:

The initial condition	   : 0 0 0 0 0 …
After the first operation  : 1 1 1 1 1 …
After the second operation : 1 0 1 0 1 …
After the third operation  : 1 0 0 0 1 …
After the fourth operation : 1 0 0 1 1 …
After the fifth operation  : 1 0 0 1 0 …

The later operations cannot change the condition of the fifth lamp any more. So the answer is 0.

[align=left]Author[/align]
LL

#include <iostream>
using namespace std;

int main()
{
int n;
while (cin >> n)
{
int num=0;
for (int i=1; i<=n; i++)
{
if (n %i ==0)
num++;
}
if (num%2 == 0)
cout << 0;
else
cout << 1;
cout <<endl;
}
return 0;
}

该题实际上就是一个求数因子个数的题目，刚开始我的第一思路就是打算用数组去保留每次的结果，但是很明显会超时。并且非常复杂，再仔细看看题目，看到了无限次，发现经过几次操作之后前面的一些数的结果不会改变了，然后仔细一想这就是一个求因子的！！