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
Consider the second test case:
Hint
hint 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 ",(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 "); 27 else printf("1 "); 28 } 29 return 0; 30 }