1049. Counting Ones (30)

最新推荐文章于 2025-09-07 17:00:17 发布

转载最新推荐文章于 2025-09-07 17:00:17 发布 · 49 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/grglym/p/7802501.html

文章标签：

#面试

本文介绍了一种高效的算法，用于计算从1到给定整数N之间所有数字中1出现的次数。通过分解数字并考虑每一位上1可能出现的情况，该算法避免了逐个检查每个数字的传统方法，显著提高了计算效率。

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:

5

题目大意：即给定一个整数N，求从1~N中包含过少个1。记得当时研究生面试时就问了这个题目，给了个整数，来求包含的1的个数。当时拿到题目，想了3分钟，面试官问答案，我还没求解出答案，只有思路。通过讲解解题思路，分别考虑1在各位、十位、百位....的情况。现在再看这题目，原来的思路还是错了。具体的思路可以参考网上博客。

#include<iostream>
#include<cstdio>
using namespace std;
int main(){
	int n;
	scanf("%d",&n);
	int high=0;
	int low=0;
	int cur=0;
	int base=1;
	int cnt=0;
	while(n/base!=0){
		high=n/(base*10);
		low=n%base;
		cur=(n/base)%10;
		switch(cur){
		case 0:
			cnt+=high*base;
			break;
		case 1:
			cnt+=high*base+low+1;
			break;
		default:
			cnt+=(high+1)*base;
		}
		base*=10;	
	}
	printf("%d\n",cnt);
	return 0;
}

转载于:https://www.cnblogs.com/grglym/p/7802501.html