1049. Counting Ones (30)【计算1 的个数】——PAT (Advanced Level) Practise

最新推荐文章于 2019-08-23 07:28:26 发布

闲云阁

最新推荐文章于 2019-08-23 07:28:26 发布

阅读量1.2k

点赞数

CC 4.0 BY-SA版权

分类专栏：浙大PAT 浙大Pat 文章标签： pat 1049

本文为博主原创文章，转载请标明出处 http://blog.youkuaiyun.com/xianyun2009

本文链接：https://blog.youkuaiyun.com/xianyun2009/article/details/50198503

浙大PAT 同时被 2 个专栏收录

106 篇文章

订阅专栏

浙大Pat

106 篇文章

订阅专栏

本文详细介绍了如何通过编程计算从1到任意正整数N之间的数字1的总数量，包括输入输出规范、解题思路及AC代码实现。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目信息

1049. Counting Ones (30)

时间限制10 ms
内存限制65536 kB
代码长度限制16000 B

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

解题思路

《编程之美》中计算0到N包含数字1的个数问题

AC代码

#include <cstdio>

int get(long long n){
    long long cur = 0, before = 0, after = 0, i = 1, cnt = 0;
    while ((n / i) != 0){
        cur = n / i % 10;
        before = n / i / 10;
        after = n - (n / i) * i;
        if (cur == 0){
            cnt += before * i;
        }else if (cur == 1){
            cnt += before * i + after + 1;
        }else {
            cnt += (before + 1) * i;
        }
        i *= 10;
    }
    return cnt;
}

int main()
{
    long long a;
    scanf("%lld", &a);
    printf("%d\n", get(a));
    return 0;
}