Making Sequences is Fun（简单枚举）

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http://codeforces.com/problemset/problem/373/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 nto 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

题意：S(n)表示一个数字的位数，w表示拥有的金钱，m表示起始数，k表示每一位数字的花费。问从m开始最多可以有多少个数。

注意：理解题意后，把过程简单化。注意 m的上限是10的16次方,m增加的上限超过10的16次方，所以初始化b数组的时候，记得弄大一些。

#include"iostream"#include"cstring"#include"cstdio"#define maxn 50005#define LL __int64using namespace std;LL w,m,k,cnt,ans;LL b[20];int main(){    b[0]=1,b[1]=10;    for(int i=2;i<=20;i++)    {        b[i]=b[i-1]*10;    }    while(~scanf("%I64d%I64d%I64d",&w,&m,&k))    {        w/=k;        ans=0;       for(int i=0;i<20;i++)            if(m<b[i])           {               cnt=i;               break;           }       while(w)       {           if(w>=(b[cnt]-m)*cnt)           {               ans += b[cnt]-m;               w -= (b[cnt]-m)*cnt;               m = b[cnt];           }           else           {               ans += w/cnt;               w=0;           }           cnt++;       }        printf("%I64d\n",ans);    }    return 0;}

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