PAT 1049. Counting Ones

最新推荐文章于 2021-07-30 10:58:10 发布

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最新推荐文章于 2021-07-30 10:58:10 发布

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【PAT 1049】 Counting Ones 数学规律

1049. Counting Ones (30)

时间限制

10 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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:

#include <iostream>
#include <cstdio>
#include <algorithm>

using namespace std;


int main()
{
    int n;
    scanf("%d", &n);

    int cnt = 0;
    int factor = 1;
    int higer, lower, cur;

    while(n / factor != 0) {
        higer = n / (factor * 10);
        lower = n - (n / factor) * factor;
        cur = (n / factor) % 10;

        switch(cur) {
    case 0:
        cnt += higer * factor;
        break;
    case 1:
        cnt += higer * factor + lower + 1;
        break;
    default:
        cnt += higer * factor + factor;
        break;
        }

        factor = factor * 10;
    }

    printf("%d", cnt);
    return 0;
}