Page number calculation

/**
* Problem: the page number of a book is encoded from the Natural Number 1 to n. Each page number has no leading 0, that is, the 6th page is encoded as 6, instead of <br>
* 06, or 006. The requirement for the number counting problem is to calculate the total page number n of the given book and the number of times the knife is used in all the pages of the book. <br>
* 0, 1 ,...... 9 .. 1 <= n <= 10 ^ 9.
* Train of Thought: For a digit t (0 <= t <= 9), consider the single digit, ten digits ...... To the highest bit of n, the number of these bits is t, and then the score is <br>
* Do not add up, that is, the number of t in all pages.
* <Br> assume that all n> = 1

**/

Class numbercount {private int N; private int length; private int [] result; Private Static void confirm (int n, int [] result) {// method used for verification, calculate the number of each integer one by one while (n> 0) {result [n % 10] ++; N/= 10 ;}} public static void test () {// test the function int [] n = {9, 23, 301,3423, 34830,000000, 83049839}; For (Int J: N) {numbercount NC = new numbercount (j ); int [] result = NC. getcount (); system. out. println ("n =" + J); For (int t = 0; t <10; t ++) {system. out. println ("digit = "+ T +" number = "+ result [T]);} int [] con = new int [10]; for (int K = 1; k <= J; k ++) Confirm (K, con); For (int K = 0; k <10; k ++) if (CON [k]! = Result [k]) {system. Out. println ("error! "); For (int t: con) system. out. println (t); break ;}} public numbercount (int n) {This. N = N; length = lengthof ();}/** get the number of each 0-9 digit in all pages */Public int [] getcount () {If (result = NULL) {result = new int [10]; Result [0] = getzerocount (); For (Int J = 1; j <10; j ++) result [J] = getcount (j);} return arrays. copyof (result, result. length);} private int getzerocount () {// calculate the number of 0 without leading 0. Therefore, int Pos = 1, TK = 10, and zeronumber = 0 must be calculated separately; while (Pos <L Ength) {// 0 cannot be in the highest bit zeronumber + = (N/tk-1) * (tk/10); // No leading 0, so reduce 1, after the number of pos-1 bit composition is tk/10if (Pos> 1) {// consider the pos-1 bit int NA = n % TK; int second = na * 10/TK; // Number of pos-1 for if (second = 0) {// The pos-1 is 0, then, the number of N/tk * TK values is Na + 1 zeronumber + = Na + 1;} else {// if this bit is not 0, then the number of all the combinations that match the pos-1 bit, that is, TK/10 zeronumber + = tk/10;} else {zeronumber + = (n <10? 0: 1);} POS ++; TK * = 10;} return zeronumber;} private int getcount (int K) {// calculate the number of 1-9 int Pos = 1, TK = 10, number = 0; while (Pos <= length) {number + = N/tk * (tk/10); If (Pos> 1) {int NA = n % TK; int second = na * 10/TK; If (second = k) {number + = Na-K * tk/10 + 1 ;} else if (second> K) {number + = tk/10;} else {number + = n % 10> = K? 1:0;} POS ++; TK * = 10;} return number;} private int lengthof () {// calculate the decimal length of N, int r = 0, T = 1; while (T <= N) {r ++; T * = 10;} return r ;}}