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杭电 2503
Switch Game
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 10590 Accepted Submission(s): 6446
Problem Description
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 ).
Input
Each test case contains only a number n ( 0< n<= 10^5) in a line.
Output
Output the condition of the n-th lamp after infinity operations ( 0 - off, 1 - on ).
Sample Input
15
Sample Output
10Consider 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.Hinthint
Author
LL
Source
校庆杯Warm Up
将步骤写出来自己观察 发现 这道题的意思是看输入的那个数
除了1之外的是否有其他的因子，要是有的话就计算因子的总数
要是总数为奇数的话说明最终加上1的那次偏转 共有偶数次偏转
开始的时候的为0即：
除一外 有奇数个因子的数或者是1的数 最终结果输出1
为偶数 的最终输出结果为零
代码如下：
#include<stdio.h>
int prime(int n)
{
int count=0,i;
if(n==1)
return 1;
for(i=2;i<=n/2;i++)
{
if(n%i==0)
count++;
}
return count;
}
int main()
{
int n;
while(~scanf("%d",&n))
{
if(prime(n)&1)
printf("1\n");
else
printf("0\n");
}
return 0;
}
int prime(int n)
{
int count=0,i;
if(n==1)
return 1;
for(i=2;i<=n/2;i++)
{
if(n%i==0)
count++;
}
return count;
}
int main()
{
int n;
while(~scanf("%d",&n))
{
if(prime(n)&1)
printf("1\n");
else
printf("0\n");
}
return 0;
}
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