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

 #include<cstdio>

 #include<iostream>

 using namespace std;

 int main(){

     long long N, a = , ans = , left, mid, right;

     scanf("%lld", &N);

     while(N / (a / ) != ){

         left = N / a;

         mid = N % a / (a / );

         right = N % (a / );

         if(mid == )

             ans += left * (a / );

         else if(mid == )

             ans += left * (a / ) + right + ;

         else if(mid > )

             ans += (left + ) * (a / );

         a *= ;

     }

     printf("%d", ans);

     cin >> N;

     return ;

 }

总结：

1、本题应寻找规律完成。对于数字abcde来说，讨论第c位为1的有多少个数字，分为左边：ab，中间c， 右边de。当c = 1时，ab可以取0到ab的任意数字，当取0到ab - 1时，de位任意取。当ab取ab时，de位只能从0 到 de， 故共有ab * 1000 + de + 1个。  当c = 0时，为了让c能 =  1， ab位只能取0 到 ab - 1共ab种可能，de位任取，故共有ab*1000种可能。  当c > 1时，ab位可以取0到ab共ab+1种可能，de任取，故共有 （ab + 1）* 1000种可能。