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
Sample Output
1 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⊙)…
1 #include<iostream>
2 #include<stdio.h>
3 #include<cstring>
4 #include<cstdlib>
5 using namespace std;
6
7 int num_Euler(int n)
8 {
9 int i;
10 int num,ans=1;
11 for(i=2;i*i<=n;i++)
12 {
13 if(n%i==0)
14 {
15 num=1;
16 while(n%i==0)
17 {
18 n=n/i;
19 num++;
20 }
21 ans=ans*num;
22 }
23 }
24 if(n!=1)
25 ans=ans*2;
26 return ans;
27 }
28 int main()
29 {
30 int n;
31 while(scanf("%d",&n)>0){
32 int m=num_Euler(n);
33 printf("%d\n",(m&1));
34 }
35 return 0;
36 }
1 #include<iostream>
2 #include<stdio.h>
3 #include<cstring>
4 #include<cstdlib>
5 using namespace std;
6
7 bool dp[100002];
8
9 void init()
10 {
11 int i,j;
12 memset(dp,true,sizeof(dp));
13 for(i=2;i<=100000;i++)
14 {
15 for(j=i;j<=100000;j=j+i)
16 if(dp[j]==true) dp[j]=false;
17 else dp[j]=true;
18 }
19 }
20 int main()
21 {
22 int n;
23 init();
24 while(scanf("%d",&n)>0)
25 {
26 if(dp[n]==false)printf("0\n");
27 else printf("1\n");
28 }
29 return 0;
30 }
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