1049 Counting Ones

最新推荐文章于 2020-03-23 15:22:45 发布

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本文深入解析了1049CountingOnes算法，该算法旨在计算从1到任意正整数N中所有数字中1的总数。通过详细分析输入输出规格，给出了一段C++实现代码，利用循环结构巧妙地解决了问题。

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1049 Counting Ones （30 point(s)）

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:

Sample Output:

想偷懒，见算法笔记p199

#include <cstdio>



int main()
{
	int i,n,sum=0,left,right,a=1;
	scanf("%d",&n);
	while(1)
	{
		i=n/a%10;
		left=n/a/10;
		right=n%a;
		if(i==0)
			sum+=(left)*a;
		else if(i==1)
			sum+=left*a+right+1;
		else
			sum+=(left+1)*a;
		a*=10;
		if(left==0)
			break;
	}
	printf("%d",sum);

}