Evaluate the number of 1 + 11 + 111 + 1111 +... + N 1 and how many 1?

Question: How many of the plus 1 + 11 + 111 + 1111 +... + N 1 and the middle 1?

Solution: Calculate from a single bit to the highest bit. For example, if the number is always increased to 2014 1, then a total of 2014 1 bits are added. First, judge the sum of the two bits (whether the single bit is 1 ), after the sum, the carry numbers are 201, of which 2, 0, and 1 correspond to the carry numbers of thousands, hundreds, and ten respectively. Then, the sum of the ten digits is 2013, and the first carry 1 is added, calculate the carry (compared to the ten-digit calculation result, only 20 in the single-digit calculation result is the carry value. In fact, 1 of the 201-digit calculation is used out of the ten-digit calculation, divide 201 by 10, which is the carry forward of the 10 digits), and so on.

The program code is as follows:

# Include <stdio. h> void sum1_compute (int n) {int sum; int carry_num = 0, Count = 0; while (n> 0) {sum = N + carry_num % 10; if (1 = sum % 10) Count ++; carry_num/= 10; carry_num + = sum/10; n --;} printf ("sum contains (1) % d. \ n ", count) ;}int main () {int N; printf (" Please input N: "); scanf (" % d ", & N ); sumatecompute (n );}

The result of running the program on Ubuntu is: