Codeforces Round #219 (Div. 2) B. Making Sequences is Fun

B. Making Sequences is Fun

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

We'll define S(n) for positive integer n as follows: the number of the n's digits in the decimal base. For example,S(893) = 3, S(114514) = 6.

You want to make a consecutive integer sequence starting from number m (m, m + 1, ...). But you need to payS(n)·k to add the number n to the sequence.

You can spend a cost up to w, and you want to make the sequence as long as possible. Write a program that tells sequence's maximum length.

Input

The first line contains three integers w (1 ≤ w ≤ 1016), m (1 ≤ m ≤ 1016), k (1 ≤ k ≤ 109).

Please, do not write the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, coutstreams or the %I64d specifier.

Output

The first line should contain a single integer — the answer to the problem.

Sample test(s)

input

9 1 1

output

9

input

77 7 7

output

7

input

114 5 14

output

6

input

1 1 2

output

0

    多么典型的二分枚举呀，不过需要注意小细节。

#include <iostream>
#include <string>
#include <string.h>
#include <map>
#include <stdio.h>
#include <algorithm>
#include <queue>
#include <vector>
#include <math.h>
#include <set>
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
using namespace std;
typedef unsigned long long LL ;

LL  dp[19] ;
LL  num[19] ;

void init(){
     num[1] = 1 ;
     for(int i = 2 ; i <= 19 ; i++)
        num[i] = num[i - 1] * 10 ;
     for(int i = 1 ; i <= 18 ; i++)
        dp[i] = i * (num[i+1] - num[i]) ;
}

LL get_all_S(LL x){
   LL sum = 0 ;
   int i ;
   for(i = 2 ; i <= 18 ; i++){
      if(x >= num[i])
        sum += dp[i-1] ;
      else
        break ;
   }
   sum += (i-1)*(x-num[i-1]+1) ;
   return sum ;
}

LL M ,W , K;

int judge(LL x){
    return W >= K*(get_all_S(x) - get_all_S(M-1)) ;
}

LL calc(){
    LL L ,R ,mid , ans = -1;
    L = M ;
    R = (LL)1000000000000000000 ;
    while(L <= R){
        mid = (L + R) >> 1 ;
        if(judge(mid)){
            ans = mid ;
            L = mid + 1 ;
        }
        else
            R = mid -1 ;
    }
    return ans==-1?0:ans - M + 1 ;
}

int main(){
    init() ;
    LL x ;
    while(cin>>W>>M>>K){
        cout<<calc()<<endl ;
    }
    return 0 ;
}