51nod1009 (1 of the number)

Title Link: https://www.51nod.com/onlineJudge/questionCode.html#!problemId=1009

Test instructions: Chinese question eh ~

Ideas: Consider each digit on the occurrence of 1 times, and then sum it up.

Note that the current digit number is 0, 1, 2~9 three cases of the counting method slightly different, think of the other details here is simple ...

Code:

1#include <bits/stdc++.h>2 #definell Long Long3 using namespacestd;4 5 intMainvoid){6ll N, ans=0, gg=1;7CIN >>N;8ll t=N;9      while(t) {Tenll cnt=t%Ten; Onell xy=n/(gg*Ten); A         if(cnt==0){//situation where the current number is 0 -ans+=xy*GG; -}Else if(cnt==1){//situation where the current number is 1 theans+=xy*GG; -ans+=n%gg+1; -}Else{ -ans+= (xy+1) *gg;//The current number is the case of 2~9 +         } -gg*=Ten; +T/=Ten; A     } atcout << ans <<Endl; -     return 0; -}

　　

51nod1009 (1 of the number)