CodeForces 373 B. Making Sequences is Fun_s, 0 9000000000000000-优快云博客

本文介绍了一种算法，用于构建从指定起点开始的最长连续整数序列，其中每个数的加入成本与其十进制位数成正比。通过合理预算管理，最大化序列长度。

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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;
}