1049. Counting Ones (30)-优快云博客

本文链接：https://blog.youkuaiyun.com/fantasydreams/article/details/79305611

本文介绍了一个计算从1到给定正整数N（N<=2^30）中所有数字中1出现次数的算法。通过分析每一位上1出现的规律，实现了高效的计算方法。

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1049. Counting Ones (30)
时间限制 100 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:
12
Sample Output:
5

#define _CRT_SECURE_NO_WARNINGS
#include <unordered_map>
#include <algorithm>
#include <iostream>
#include <iomanip>
#include <cstring>
#include <string>
#include <vector>
#include <stack>
#include <queue>
#include <map>

using namespace std;

int main()
{
#ifdef _DEBUG
    freopen("data.txt", "r+", stdin);
#endif // _DEBUG

    long long c = 0;
    int num, bit = 1, left, right,curb;
    cin >> num;
    while (num / bit)
    {
        left = num / bit / 10;
        right = num % bit;
        curb = num / bit % 10;
        if (curb == 0)
            c += left * bit;
        else if (curb == 1)
            c += left * bit + right + 1;
        else
            c += (left + 1) * bit;

        bit *= 10;
    }
    cout << c << endl;
    return 0;
}