Codeforce Round #219 Div2 B

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 pay S(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, cout streams 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

Input

77 7 7

Output

Input

114 5 14

Output

Input

1 1 2

Output

 1 #include <cstdio>
 2 #include <vector>
 3 #include <cmath>
 4 #include <queue>
 5 #include <cstring>
 6 #include <iostream>
 7 #include <algorithm>
 8 using namespace std;
 9 #define INF 0x7fffffff
10 #define mod 1000000007
11 #define ll unsigned long long int
12 #define maxn 1005
13 ll w, k, m, ans;
14 int dp[maxn][maxn];
15 int a[101000];
16 char s[101000];
17 ll slen(ll x){
18     ll n = x;
19     ll t = 0;
20     ll i = 0;
21     while (n){
22         n /= 10;
23         i++;
24     }
25     ll xx = 9;
26     ll j = 1;
27     for (j = 1; j < i; j++){
28         t += xx*j;
29         xx *= 10;
30     }
31     xx = xx / 9;
32     t += (x - xx + 1)*j;
33     return t;
34 }
35 ll fun(){
36     ll l = m, r = 10000000000000000+m, mid;
37     while (l < r){ 
38         mid = (l + r) >> 1;
39         if (k*(slen(mid) - slen(m - 1)) > w)r = mid;
40         else l = mid + 1;
41     }
42     return l - m;
43 }
44 int main(){
45     cin >> w >> m >> k;
46     ans = fun();
47     cout << ans;
48     return 0;
49 }

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posted @ 2013-12-14 00:06 HaibaraAi 阅读(152) 评论(0) 收藏举报

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