# Ural 1209. 1, 10, 100, 1000... 一道有趣的题

## 1209. 1, 10, 100, 1000...

Time limit: 1.0 second
Memory limit: 64 MB
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 integer N in the first line (1 ≤ N ≤ 65535). The i-th of N left lines contains the integer Ki — the number of position in the sequence (1 ≤ Ki ≤ 231 − 1).

### Output

You are to output N digits 0 or 1 separated with a space. More precisely, the i-th digit of output is to be equal to the Ki-th digit of described above sequence.

### Sample

inputoutput
4
3
14
7
6

0 0 1 0

Problem Author: Alexey Lakhtin
Problem Source: USU Open Collegiate Programming Contest October'2002 Junior Session

有一个奇怪的排列：$110100100010000\cdots$这样。问这个序列第$K$位是$0$还是$1$。

（听说有人拿二分来做？听说有人直接$O(N)$？咦？这不是简单算术题吗！）（逃）
仔细观察发现$1$的出现序号是$1,2,4,7,11,16,22\cdots$这样。你以为我接下来要说找规律吗？naive。
再仔细观察发现就是$x*(x+1)/2+1=K(x\geq0)$，这还能忍？化简移项得$x^2+x+2-2K=0$，$b^2-4ac=8K-7$，求根公式$\frac{-1\pm\sqrt{8K-7}}{2}$，那么，是要算一下求根公式的值看是不是整数吗？我们发现$8K-7$是一个奇数！它若能开平方出一个整数，那也是一个奇整数！因此只需要判断$\sqrt{8K-7}$是不是整数就好了。

 1 #include<stdio.h>
2 #include<math.h>
3
4 int main()
5 {
6     int n;
7     double k, d;
8     scanf("%d", &n);
9     while(n--) {
10         scanf("%lf", &k);
11         k = 8*k-7; d = sqrt(k);
12         printf("%d%c", d==floor(d), n?' ':'\n');
13     }
14     return 0;
15 }

posted @ 2016-03-25 20:14  BlackStorm  阅读(604)  评论(0编辑  收藏