Timus 1209. 1, 10, 100, 1000... 根据数列推导公式-优快云博客

本文介绍了一种快速求解特定序列中任意位置数字的方法。针对由递增的10的幂次组成的序列，通过数学公式推导而非循环迭代的方式确定序列任一位置上的数字。示例程序展示了如何高效计算并输出指定位置的数字。

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Let's consider an infinite sequence of digits constructed of ascending powers of 10 written one after another. Here is the beginning of the sequence: 110100100010000… You are to find out what digit is located at the definite position of the sequence.

Input

There is the only integerNin the first line (1 ≤N≤ 65535). Thei-th ofNleft lines contains the integerKi— the number of position in the sequence(1 ≤Ki≤ 231− 1).

Output

You are to outputNdigits 0 or 1 separated with a space. More precisely, thei-th digit of output is to be equal to theKi-th digit of described above sequence.

Sample

input	output
4 3 14 7 6	0 0 1 0

这道题的关键就是要推导出公式了，如果使用循环，那么肯定是超时的。

根据数列的特征，知道循环的周期是1， 2， 3， 4， 5……

那么就可以知道需要计算k的位置是到了那个周期了，提示到这，看程序吧

void sequence1101001000()
{
	int T = 0;
	long long k = 0, n = 0;
	cin>>T;
	while (T--)
	{
		cin>>k;
		if (k < 3) 
		{
			cout<<1<<' ';
			continue;
		}
		n = sqrt((double)(k<<1));
		n = ((1+n)*n)>>1;
		if (n+1 == k) cout<<1<<' ';
		else cout<<0<<' ';
	}
}