How many 0's?

 

Time Limit: 3sec    Memory Limit:32MB

Description

A Benedict monk No. 16 writes down the decimal representations of all natural numbers between and including m and n, m ≤ n. How many 0's will he write down?

Input

Input consists of a sequence of lines. Each line contains two unsigned 32-bit integers m and n, m ≤ n. The last line of input has the value of m negative and this line should not be processed.

Output

For each line of input print one line of output with one integer number giving the number of 0's written down by the monk.

Sample Input

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10 11100 2000 5001234567890 23456789010 4294967295-1 -1

Sample Output

122929876543043825876150

// Problem#: 1370// Submission#: 2196597// The source code is licensed under Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License// URI: http://creativecommons.org/licenses/by-nc-sa/3.0/// All Copyright reserved by Informatic Lab of Sun Yat-sen University#include<stdio.h>#include<iostream>using namespace std;long long get_ans(long long u){    long long tmp=1,ans=1,k=u/10;        while (k)    {        ans+=(k-1)*tmp;        if (u-tmp*k*10+1>=tmp)            ans+=tmp;        else            ans+=u-tmp*k*10+1;        tmp*=10;        k/=10;    }    return ans;}// 截断，先算出来从1----n   1----m-1   之后就可以知道一共有多少个0了 int main(){    long long ans,m,n;    while (1)    {        cin>>m>>n;        if (m<0) break;                ans=get_ans(n);        if (m>0) ans-=get_ans(m-1);         cout<<ans<<endl;    }    return 0;}                                 

#include <iostream>using namespace std;long long pow10[]={1, 10, 100, 1000, 10000, 100000, 1000000, 10000000, 100000000, 1000000000};long long num0(long long n){     long long i;     long long cnt=1+n/10;     if(n<0)          return 0;     for(i=1; pow10[i]<=n/10; i++)     {          long long left=(n/10)/pow10[i];         long long mid =( n/pow10[i] )%10;         long long rit =n%pow10[i];         if(mid==0)             cnt+=(left-1)*pow10[i] + rit+1;         else             cnt+=left*pow10[i];     }     return cnt;}int main(){      long long m, n;      while(cin>>m>>n && m>=0){          cout<<num0(n)-num0(m-1)             <<endl;     }      return 0;}