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  • Switch Game :因子数

    A - Switch Game

    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

    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.

    题目描述:输入一个n,取i从1到n,在1~n中,是i的倍数,值就变一次,求最后变换完后第n个数的值。

    No.1:查找含有多少因子

    #include<iostream>
    #include<algorithm>
    using namespace std;
    
    int main() {
    	int n,ans;
    	while (cin >> n) {
    		ans = 2;
    		if (n <= 3)cout<<1<<endl;
    		else {
    		for (int i = 2; i*i<= n; i++) {		//枚举含有几个因子 
    			if (n%i == 0) {
    				if (i*i == n)ans++;			//
    				else ans += 2;				//如果不是i*i,那除了i是因子,n/i也是,所以上边枚举到i*i就可以了。 
    			}
    		}
    		if (ans & 1)cout<<1<<endl;
    		else cout<<0<<endl;
    		}
    	}
    	return 0;
    }
    

     No.2:写博客时刚想到,可以直接判断n是不是某个整数的平方,是的话,因子肯定为奇数,所以结果为1.,不是结果为0

    #include<iostream>
    #include<algorithm>
    using namespace std;
    
    int main() {
    	double n,ans;
    	while (cin >> n) {
    		int ans=0;
    		for(int i=1;i<400;i++){            //400*400就大于1e5了
    			if(i*i==n)ans=1;
    		}
    		if(ans)cout<<"1
    ";
    		else cout<<"0
    ";
    	}
    	return 0;
    }    
    



     
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  • 原文地址:https://www.cnblogs.com/52dxer/p/10548007.html
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