Switch Game

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10200    Accepted Submission(s): 6175

Problem DescriptionThere 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 ). InputEach test case contains only a number n ( 0< n<= 10^5) in a line. OutputOutput the condition of the n-th lamp after infinity operations ( 0 – off, 1 – on ). Sample Input1
5 Sample Output1
0

Hint

hint

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. AuthorLL  转化求n个因子个数，暴力0ms过了(⊙o⊙)…

 #include<iostream>
 #include<stdio.h>
 #include<cstring>
 #include<cstdlib>
 using namespace std; int num_Euler(int n)
 {
     int i;
     int num,ans=;
     for(i=;i*i<=n;i++)
     {
         if(n%i==)
         {
             num=;
             while(n%i==)
             {
                 n=n/i;
                 num++;
             }
             ans=ans*num;
         }
     }
     if(n!=)
         ans=ans*;
     return ans;
 }
 int main()
 {
     int n;
     while(scanf("%d",&n)>){
         int m=num_Euler(n);
         printf("%d\n",(m&));
     }
     return ;
 }

 #include<iostream>
 #include<stdio.h>
 #include<cstring>
 #include<cstdlib>
 using namespace std; bool dp[]; void init()
 {
     int i,j;
     memset(dp,true,sizeof(dp));
     for(i=;i<=;i++)
     {
         for(j=i;j<=;j=j+i)
             if(dp[j]==true) dp[j]=false;
             else dp[j]=true;
     }
 }
 int main()
 {
     int n;
     init();
     while(scanf("%d",&n)>)
     {
         if(dp[n]==false)printf("0\n");
         else printf("1\n");
     }
     return ;
 }