Description

A single positive integer i is given. Write a PRogram to find the digit located in the position i in the sequence of number groups S1S2…Sk. Each group Sk consists of a sequence of positive integer numbers ranging from 1 to k, written one after another. For example, the first 80 digits of the sequence are as follows: 11212312341234512345612345671234567812345678912345678910123456789101112345678910 Input

The first line of the input file contains a single integer t (1 ≤ t ≤ 10), the number of test cases, followed by one line for each test case. The line for a test case contains the single integer i (1 ≤ i ≤ 2147483647) Output

There should be one output line per test case containing the digit located in the position i. Sample Input

2 8 3 Sample Output

2 2

題目大意 數(shù)字序列按照 1 12 123 1234 12345 123456 1234567 12345678 123456789 12345678910 1234567891011…的規(guī)律進(jìn)行排列，輸入一個(gè)數(shù) n ，輸出在序列中第 n 個(gè)位置的數(shù)字。

解題思路 這是第三次做這個(gè)題目了 好像請(qǐng)教了大神兩次 這一次終于敲出來(lái)了 想問(wèn)問(wèn)自己當(dāng)初真的有那么難嗎[廢話(huà)好多haha] 1、準(zhǔn)備階段：對(duì)于每一個(gè)數(shù)字 i ，它所占的位數(shù)為 log10(i)+1，由此來(lái)進(jìn)行打表，每段序列（如題目大意中所示，空格隔開(kāi)的即為一段，序列尾的數(shù)字即為該段的標(biāo)志數(shù)）的長(zhǎng)度記為 a[i] ,截止到當(dāng)前的序列的總長(zhǎng)度即為 s[i] ；則可得a[i]=a[i-1]+log10(n)+1，s[i]=s[i-1]+a[i]。 2、求解階段：對(duì)于給定的數(shù) n ，先根據(jù) s[i] 找到第 n 個(gè)數(shù)所處的段序列，則 n-s[i-1] 即為第 n 個(gè)數(shù)在該段序列中的位置，記為st；接下來(lái)用 len 標(biāo)記每到一個(gè)數(shù)當(dāng)前序列的長(zhǎng)度，結(jié)果必滿(mǎn)足 len>=st，len-st 即為第 n 個(gè)數(shù)在最終數(shù)中的位置（由右至左從0開(kāi)始）。如：i-1=123， len-st=2，則 123/100%10=1；

代碼實(shí)現(xiàn)