Offer prove safety [31], (the number of occurrences n is an integer from 1 to 1) times an integer of 1 appearing in

Title Description

Calculated the number of integer of 1 to 13 1 arise, and calculates the number of times an integer from 100 to 1300 in 1 occur? To this end he especially counted about 1 to 13 contains the digits 1 has 1,10,11,12,13 therefore appear a total of six times, but for the problem behind him Meizhe. ACMer hope you will help him, and the problem is more generalized, the number of non-negative integer in the range 1 appears (the number of occurrences from 1 to n. 1) can quickly find any.

Solving a problem: induction

1  / ** 
2  number 1 appears on the * bits are: n-/ 10 * 1+ (n-10 = 0% 1:!? 0)
 . 3  the number appearing on ten * 1: (n / 100 ) * 10 + (IF (K>. 19) 10 the else IF (K <10) the else K 0 - 10 + 1)
 . 4  * 1 appears on the number of one hundred of: (n / 1000) * 100 + (if ( K> 199) 100 the else IF (K <100) the else 0 K - 1 + 100)
 . 5  * i that is calculated where the bits 1, i = 1, number 1 indicates a calculated number of bits, 10 denotes a computing ten the number of bits 1
 . 6  *% n-K = (I * 10)
 . 7  * COUNT (I) = (n-/ (10 * I)) + I * (IF (K> I * 2 - 1) I the else IF (K <I) 0 the else k - i + 1)
 . 8  * can be simplified so that the second half, we do not calculate i * 2 - 1, we only need to ensure k - i + 1 in the [0, i ] on the line interval, the last half of this can be written
 . 9  * min (max (I + K-1,0), I) => min (max ((n-MOD (I 10 *)) -. 1 + I, 0), I)
 10   * / 
. 11  public  static  int NumberOf1Between1AndN_Solution(int n) {
12         if(n <= 0) {
13             return 0;
14         }
15         int count = 0;
16         for(long i = 1; i <= n; i *= 10){
17             long diviver = i * 10;
18             count += (n / diviver) * i + Math.min(Math.max(n % diviver - i + 1, 0), i);
19         }
20         return count;

Solution two questions: StringBuffer ()

 1 public static int NumberOf1Between1AndN_Solution01(int n) {
 2         int count=0;
 3         StringBuffer s=new StringBuffer();
 4         for(int i=1;i<n+1;i++){
 5             s.append(i);
 6         }
 7         String str=s.toString();
 8         for(int i=0;i<str.length();i++){
 9             if(str.charAt(i)=='1'){
10                 count++;
11             }
12         }
13         return count;
14     }

Problem solution three: judge suffix

1  / * 
2       * n, only the m-th bit is to be calculated when a suffix is a suffix calculated (n / m% 10 == 1) * (B + 1),
 . 3       * i.e. (n / m% 10 == 1) determining whether the m-th bit is 1, if it is 1, then the plus (b + 1), if 1, only the prefix is calculated.
4       * (2 if the number of calculations can be changed (n / m% 10 == 2) * (B +. 1),
 . 5       * 3 if the number of calculations can be changed to (n / m% 10 == 3) * (b + 1) ... and so on)
 . 6       * / 
. 7  public  static  int NumberOf1Between1AndN_Solution02 ( int n-) {
 . 8          int ones = 0 ;
 . 9          for ( int m =. 1; m <= n-; m * = 10 ) {
 10              // one by one to obtain the values of m bits each 
. 11              int a = n-/ m;
 12 is              int B = n-% m;
13             if(a % 10 == 1){
14                 ones += (a + 8) / 10 * m +b + 1;
15             }else {
16                 ones += (a + 8) / 10 * m;
17             }
18         }
19         return ones;
20     }

test:

1 public static void main(String[] args) {
2         int n=13;
3         int number = NumberOf1Between1AndN_Solution(n);
4         System.out.println(number);
5     }
6 输出：6