1049. Counting Ones/整数中1出现的次数（从1到n整数中1出现的次数）

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

题目描述

求出1~13的整数中1出现的次数,并算出100~1300的整数中1出现的次数？为此他特别数了一下1~13中包含1的数字有1、10、11、12、13因此共出现6次,但是对于后面问题他就没辙了。ACMer希望你们帮帮他,并把问题更加普遍化,可以很快的求出任意非负整数区间中1出现的次数。

 

#include<iostream>
#include<string>
#include<sstream>
#include<math.h>
using namespace std;
int CountOne(std::string num)
{
    int len = num.length();
    if(len == 0)
        return 0;
    int fir = num[0] - '0';

    if(len == 1 && fir == 0)
        return 0;
    if (len == 1 && fir >= 1)
        return 1;

    int num1 = 0,num2,num3;
    if (fir == 1)
    {
        string re = num ,tem;
        num.erase(num.begin());
        tem = num;
        num = re;
        re = tem;
        stringstream ss;
        ss << re;
        ss >> num1;
        ++ num1;
    }
    else if(fir > 1)
    {
        num1 = pow((double)10,len - 1);
    }
    num2 = (len -1) * fir * pow((double)10,len-2);
    string tnum = num;
    tnum.erase(tnum.begin());
    num3 = CountOne(tnum);
    return num1 + num2 + num3;
}
int main()
{
    string num;
    cin >> num;
    printf("%d\n",CountOne(num));
    return 0;
}