PAT 1049 Counting Ones(数学规律)

最新推荐文章于 2024-02-16 17:47:05 发布

漂流瓶终结者

最新推荐文章于 2024-02-16 17:47:05 发布

阅读量213

点赞数

CC 4.0 BY-SA版权

分类专栏：数据结构

本文链接：https://blog.youkuaiyun.com/ZCMU_2024/article/details/84961044

数据结构专栏收录该内容

46 篇文章

订阅专栏

本文介绍了一种高效算法，用于计算从1到任意正整数N之间所有整数中数字1出现的总次数。通过巧妙地分解数字，算法避免了逐个数字检查的传统方法，显著提高了计算效率。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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:

题解

#include<stdio.h>
int CountOnes(int n)
{
    int count=0;
    int factor=1;
    int higher,lower,cur;
    while(n/factor!=0)
    {
        higher=n/(factor*10);
        lower=n-(n/factor)*factor;
        cur=(n/factor)%10;
        switch(cur)
        {
           case 0:
              count+=higher*factor;
              break;
           case 1:
              count+=higher*factor+lower+1;
              break;
           default:
              count+=higher*factor+factor;
        }
        factor=factor * 10;
    }
    return count;
}

int main()
{
    int n;
    scanf("%d",&n);
    printf("%d",CountOnes(n));
    return 0;
}