1 ~ N direct 1 appears

Refer to previous statistical ideas: count the number of times that the nth digit appears on the nth digit, 10, and, respectively.

For example, ABCDE, when calculating the number of occurrences of D-bit 1, use d as the delimiter, ABC is before, and E is after.

Considerations: (n is the length-1 of D)

When D = 0, Count = before * 10 ^ N;

When d = 1, Count = before * 10 ^ N + after;

When D> 1, Count = (before + 1) * 10 ^ N;

For example:

19x8

Count the number of times 1 on X:

1) x = 0, that is, the number of 1908 X as 1 has ~ 181x, x 0 ~ 9, 19 is before, and 8 is after

Count = 19*10 ^ 1;

2) x = 1, that is, the number of 1918 X as 1 has ~ 181x, x 0 ~ 9; in addition, 1910 ~ 1918, then 19 is before, 8 is after

In this case, Count = 19*10 ^ 1 + (8 + 1 );

3) x> 1, for example, the number of 1928 X as 1 has ~ X, 0 ~ 9, 19 is before, and 8 is after

In this case, Count = (19 + 1) * 10 ^ 1;

In particular, when X is at the leftmost end, before is 0, and after is 0 at the rightmost end.

#include <stdio.h>int Count1(int n){    int count = 0,//1出现总次数    bitCount = 0,//某位1出现次数    base = 1,//基数    before = n,after = 0,  //从最右开始，则Before = n，After = 0    bitN = 0;//第N位数    while(before)//向左移，还有数时循环    {        after = n % base;        before = n / (base * 10);        bitN = (n / base) % 10;        if(bitN > 1)        {            bitCount = (before + 1) * base;        }        else if(bitN == 0)        {            bitCount = (before) * base;        }        else        {            bitCount = (before) * base + (after + 1);        }        base *= 10;        count += bitCount;    }    return count;}int main() {   int n = 121;   printf("%d\n",Count1(n));   return 0;}

 

1 ~ N direct 1 appears