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 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
1
5
5
Sample Output
1
0
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.
Author
LL
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 ;
}