1049. Counting Ones (30)

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

特别经典的算法，借鉴学习了

#include <iostream>
#include <map>
#include <queue>
#include <stdio.h>
#include <vector>
#include <algorithm>
#include <stack>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <string>
using namespace std;


int c,base;
int main()
{	
	int	n;
	while(scanf("%d",&n)!=EOF){
		c=0;base=1;     
		while(n/base!=0){
			int low=n%base;         //当前权值的低位
			int high=(n/base)/10;   //当前权值的高位
			int now=(n/base)%10;    //当前权值位
			if(now==0)
				c+=high*base;
			else if(now==1)
				c+=high*base+low+1;
			else
				c+=(high+1)*base;
			base*=10;    //注意
		}
		printf("%d\n",c);
	}
    return 0;
}