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  • acm日常CodeForces

    CodeForces - 318A

    Being a nonconformist, Volodya is displeased with the current state of things, particularly with the order of natural numbers (natural number is positive integer number). He is determined to rearrange them. But there are too many natural numbers, so Volodya decided to start with the first n. He writes down the following sequence of numbers: firstly all odd integers from 1 to n (in ascending order), then all even integers from 1 to n (also in ascending order). Help our hero to find out which number will stand at the position number k.

    Input
    The only line of input contains integers n and k (1 ≤ k ≤ n ≤ 1012).

    Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.

    Output
    Print the number that will stand at the position number k after Volodya’s manipulations.

    Examples
    Input
    10 3
    Output
    5
    Input
    7 7
    Output
    6
    Note
    In the first sample Volodya’s sequence will look like this: {1, 3, 5, 7, 9, 2, 4, 6, 8, 10}. The third place in the sequence is therefore occupied by the number 5.

    解题思想:
    ①开始的错误思想:列原题数组,再进行检测。×
    ②应该是观察发现,其实就是由两个等差数列组成,找规律,发现数学表达式。√

    ac代码

    #include<iostream>
    using namespace std;
    
    int main()
    {
    	long long n, k;
    	cin >> n >> k;
    	if (n % 2)
    	{
    		if (k <= n / 2 + 1)//奇数
    			cout << 2 * k - 1 << endl;
    		else
    			cout << 2 * (k-n/2-1)<< endl;
    	}
    	else
    	{
    		if (k <= n / 2)
    			cout << 2 * k - 1 << endl;
    		else
    			cout << 2 * (k - n / 2)  << endl;
    	}
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/gidear/p/10433312.html
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