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 == )
         return ;
     int fir = num[] - '';

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

     int num1 = ,num2,num3;
     if (fir == )
     {
         string re = num ,tem;
         num.erase(num.begin());
         tem = num;
         num = re;
         re = tem;
         stringstream ss;
         ss << re;
         ss >> num1;
         ++ num1;
     }
     else if(fir > )
     {
         num1 = pow((double),len - );
     }
     num2 = (len -) * fir * pow((double),len-);
     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 ;
 }