Pat(Advanced Level)Practice--1049(Counting Ones)

Pat1049代码

题目描述：

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

AC代码：

#include<cstdio>

using namespace std;

int CountOneNum(int n)
{
	int ret=0;
	int factor=1;
	int low=0;
	int cur=0;
	int high=0;
	while(n/factor!=0)
	{
		low=n-(n/factor)*factor;
		cur=(n/factor)%10;
		high=n/(factor*10);
		switch (cur)
		{
			case 0:
				ret+=high*factor;
				break;
			case 1:
				ret+=high*factor+low+1;
				break;
			default:
				ret+=(high+1)*factor;
				break;
		}
		factor*=10;
	}
	return ret;
}

int main(int argc,char *argv[])
{
	int n;
	scanf("%d",&n);
	printf("%d\n",CountOneNum(n));

	return 0;
}

更详细的分析，请看博客链接：http://blog.csdn.net/cstopcoder/article/details/22760793