Evaluate the number where 1 appears in the numbers from 1 to n

Method 1: Enumerate all numbers from 1 to n. Each number calculates the number of 1 s. The result is obtained after the sum. The complexity is O (n * lnn)

Method 2: calculate the total number of occurrences by the number of occurrences of the last bit. For example, 512,

The numbers shown in 1 include:

1 (1)

11, 21, 31,... 91 (9)

101,121,131,... 191 (10)

201,211,221 .. 291 (10)

301,311,321,... 391 (10)

401,411,421,... 491 (10)

501,511, (2)

You contributed 52 (51 + 1) X 1

The number of contributions of 10 to 1 is:

10, 11, 12,... 19 (10)

110,111,... 119 (10)

210,211 .. 219 (10)

310,311,... 319 (10)

410,411,... 419 (10)

510,511,512

(5) * 10 + 2 + 1

The number of contributions of one hundred bits is:

100, 199 (100)

(0 + 1) x 100

Through continuous analysis, we can find a rule:

If the P bit is 0, the contribution of this bit is a high number * 10 ^ P;

If P is 1, the result of contribution of this bit is a high number * 10 ^ P + low number + 1

If the P bit is greater than 1, the result of this bit contribution is (high digit + 1) * 10 ^ P

ThisAlgorithmIt is only related to the length of the input number. The length of the input number is lnn, so the complexity is lnn.

Based on the above analysisProgramAs follows:

# Include <stdio. h> int count1 (INT value) {int num = 0; while (value) {If (Value % 10 = 1) {num ++ ;} value = value/10;} return num;} int power (INT Radix, int exp) {int result = 1; for (INT I = 0; I <exp; ++ I) {result * = Radix;} return result;} int count1_1 (INT value) {int current_number = 0; int higher_parts = 0; int lower_parts = 0; int Radix = 1; int num = 0; while (value/Radix) {current_number = (value/Radix) % 10; lower_parts = value-(value/Radix) * Radix; higher_parts = value/(Radix * 10); Switch (current_number) {Case 0: num + = higher_parts * Radix; break; Case 1: num + = higher_parts * Radix + lower_parts + 1; break; default: num + = (higher_parts + 1) * Radix;} Radix * = 10;} return num ;} int main (INT argc, char ** argv) {const int kvalue = 512; int num = 0; For (INT I = 0; I <= kvalue; ++ I) {num + = count1 (I);} printf ("% d \ n", num); printf ("% d \ n", count1_1 (kvalue ));}

References:

Http://www.nowamagic.net/algorithm/algorithm_CountOccurrencesOfOne.php

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