HDU 2053 - Switch Game (開燈問題)
Switch Game
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 21842 Accepted Submission(s): 13386
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.
題意:
n盞開始為關閉的燈,經過n次操作,問第n盞燈最後的狀態。
思路:
一開始想著模擬,但是又一想一定會超時,然後就發現第n盞燈的開關和它約數的個數有關,約數的個數為偶數是燈為關閉的,約束的個數為奇數燈為開的。
程式碼:
#include<stdio.h> #include<math.h> int main() { int i,n,c; int a[100000]; while(scanf("%d",&n)!=EOF) { c=0; for(i=1;i<=n;i++) { if(n%i==0) c++; } if(c%2==0) printf("0\n"); else printf("1\n"); } return 0; }