CodeForces 373 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 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 %I64dspecifier.

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

#include <iostream>#include <cstring>#include <cstdio>using namespace std;typedef unsigned long long int LL;LL sates[25]={0,10,90,900,9000,90000,900000,9000000,90000000,900000000,9000000000,90000000000,900000000000,9000000000000,90000000000000,900000000000000,9000000000000000,90000000000000000,900000000000000000,9000000000000000000};LL w,m,k;int getlen(LL x){    int cnt=0;    while(x) cnt++,x/=10;    return cnt;}LL E(int x){    LL e=1;    while(x) e*=10,x--;    return e;}int main(){    cin>>w>>m>>k;    int len=getlen(m),pos=len;    if(w<k*len) {puts("0");return 0;}    LL res=E(len)-m;    if(w>res*k*len)    {        w-=res*k*pos;        LL W=w;        while(W>0)        {            w=W;            pos++;            if(W>k*pos*sates[pos]) W-=k*pos*sates[pos];            else break;        }    }    res=w;    LL low=E(pos-1),mid;    if(pos==len) low=m;    mid=low+res/(k*pos)-1;    cout<<mid-m+1<<endl;    return 0;}