CodeForces 373B——模拟——Making Sequences is Fun

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 numbern 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 Input

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<cstdio>

#include<cstring>

#include<algorithm>

using namespace std;

long long  w, m , k;

long long temp1(long long x)

{

    long long y = 0;

    while(x > 0){

        x /= 10;

        y++;

    }

    return y;

}



long long temp2(long long x)

{

    long long num = 1;

    for(int i = 1; i <= x; i++)

        num *= 10;

    return num;

}



int main()

{

    long long sum;

   while(~scanf("%lld%lld%lld", &w, &m, &k)){

    long long a = temp1(m);//位数

    long long t = w / k;//需要减掉的数

    long long t1 = t / a;//所需要的数的个数

    long long t2 = temp2(a);//10.....

    long long ans = 0;

    while(t > 0){

        if(t1 > t2 - m){

            t -= (t2 - m) * a;

            ans += t2 - m;

           // printf("%d\n",ans);

            a++;

            t1= t / a;

            t2 = temp2(a);

            m = temp2(a-1);

        }

        else {

            t -= t1 * a;

            ans += t1;

            break;

        }

    }

    printf("%lld\n", ans);

    }

    return 0;

}