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;}