CodeForces 373 B. Making Sequences is Fun

最新推荐文章于 2020-11-05 16:00:45 发布

最新推荐文章于 2020-11-05 16:00:45 发布 · 148 阅读

本文介绍了一种算法，用于在给定成本限制下构建最长的连续整数序列。该算法考虑了从特定数字开始，每增加一个数字所需的成本，并通过优化求解最大长度。

We'll defineS(n)for positive integernas follows: the number of then's digits in the decimal base. For example,S(893) = 3,S(114514) = 6.

You want to make a consecutive integer sequence starting from numberm(m, m + 1, ...). But you need to payS(n)·kto add the numbernto the sequence.

You can spend a cost up tow, 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 integersw(1 ≤ w ≤ 1016),m(1 ≤ m ≤ 1016),k(1 ≤ k ≤ 109).

Please, do not write the%lldspecifier to read or write 64-bit integers in C++. It is preferred to use thecin,coutstreams 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;
}