PAT-A 1049. Counting Ones

The task is simple: given any positive integer N, you are supposed to count the total number of 1’s in the decimal form of the integers from 1 to N. For example, given N being 12, there are five 1’s in 1, 10, 11, and 12.

Input Specification:

Each input file contains one test case which gives the positive N (<=230).

Output Specification:

For each test case, print the number of 1’s in one line.

Sample Input:

12

Sample Output:

5

思路参考：
http://blog.csdn.net/tiantangrenjian/article/details/19908885

程序代码：

#include<stdio.h>
#include<math.h>
#include<string.h>
char c[11]={0};
int count_1(int n);
int main()
{
    int n,i=0;
    scanf("%d",&n);
    int sum = count(n);
    printf("%d",sum);
    return 0;
}
int count(int n)
{
    int factor = 1;
    int sum = 0,lower,high,cur;
    while(n/factor!=0)
    {
        lower = n%(factor);
        high = n/(factor*10);
        cur = (n/factor)%10;
        switch(cur)
        {
        case 0:
            sum += high *factor;
            break;
        case 1:
            sum += high *factor + lower+1;
            break;
        default:
            sum += (high+1)*factor;
        break;
        }
        factor=factor*10;
    }
    return sum;
}