Switch Game
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 12748 Accepted Submission(s): 7753
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 0Consider 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.Hinthint
Author
LL
#include <iostream> using namespace std; int main() { int n; while (cin >> n) { int num=0; for (int i=1; i<=n; i++) { if (n %i ==0) num++; } if (num%2 == 0) cout << 0; else cout << 1; cout <<endl; } return 0; }
标题实际上是一些必要知识的数量因素,刚开始我的第一个想法是希望保留各自为阵的结果,但很明显的超时。而且很复杂,再详细看题目,见无限,经过一些数字前面发现了几个经营业绩将不会改变,后来我想,这是一个细致的需求因素!!