1049 Counting Ones (30)

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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 (<=2^30^).

Output Specification:

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

Sample Input:

12

Sample Output:

5

思路：

给出一个数字n，求1~n的所有数字里面出现1的个数

设当前处理至第k位，记left为第k位的高位所表示的数，now为第k位的数，right为第k位的低位所表示的数。

1. 若now==0，则ans+=left*a;
2. 若now==1，则ans+=left*a+right+1
3. 若now>=2，则ans+=（left+1）*a

C++：

#include<cstdio>
int main(){
	int n,a=1,ans=0;
	int left,now,right;
	scanf("%d",&n);
	while(n/a!=0){
		left=n/(a*10);
		now=n/a%10;
		right=n%a;
		if(now==0)ans+=left*a;
		else if(now==1)ans+=left*a+right+1;
		else ans+=(left+1)*a;
		a*=10;
	}
	printf("%d\n",ans);
	return 0;
}