Buy an Integer——UPC

题目描述

Takahashi has come to an integer shop to buy an integer.
The shop sells the integers from 1 through 109. The integer N is sold for A×N+B×d(N) yen (the currency of Japan), where d(N) is the number of digits in the decimal notation of N.
Find the largest integer that Takahashi can buy when he has X yen. If no integer can be bought, print 0.

Constraints
·All values in input are integers.
·1≤A≤10⁹
·1≤B≤10⁹
·1≤X≤10¹⁸

输入

Input is given from Standard Input in the following format:

A B X

输出

Print the greatest integer that Takahashi can buy. If no integer can be bought, print 0.

样例输入

【样例1】
10 7 100
【样例2】
2 1 100000000000
【样例3】
1000000000 1000000000 100
【样例4】
1234 56789 314159265

样例输出

【样例1】
9
【样例2】
1000000000
【样例3】
0
【样例4】
254309

提示

样例1解释
The integer 9 is sold for 10×9+7×1=97 yen, and this is the greatest integer that can be bought. Some of the other integers are sold for the following prices:
·10:10×10+7×2=114 yen
·100:10×100+7×3=1021 yen
·12345:10×12345+7×5=123485 yen
样例2解释
He can buy the largest integer that is sold. Note that input may not fit into a 32-bit integer type.

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#pragma GCC optimize("Ofast,unroll-loops,no-stack-protector,fast-math")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize (2)
#pragma G++ optimize (2)
#include <bits/stdc++.h>
#include <algorithm>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>
using namespace std;
#define wuyt main
typedef long long ll;
#define HEAP(...) priority_queue<__VA_ARGS__ >
#define heap(...) priority_queue<__VA_ARGS__,vector<__VA_ARGS__ >,greater<__VA_ARGS__ > >
template<class T> inline T min(T &x,const T &y){return x>y?y:x;}
template<class T> inline T max(T &x,const T &y){return x<y?y:x;}
//#define getchar()(p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++)
//char buf[(1 << 21) + 1], *p1 = buf, *p2 = buf;
ll read(){ll c = getchar(),Nig = 1,x = 0;while(!isdigit(c) && c!='-')c = getchar();
if(c == '-')Nig = -1,c = getchar();
while(isdigit(c))x = ((x<<1) + (x<<3)) + (c^'0'),c = getchar();
return Nig*x;}
#define read read()
const ll inf = 1e15;
const int maxn = 2e5 + 7;
const int mod = 1e9 + 7;
#define start int wuyt()
#define end return 0
start{
    ll a=read,b=read,c=read;
    ll left=0,right=1000000001;
    while(right-left>1){
        ll temp=(left+right)>>1;
        ll temp2=temp;
        int cnt=0;
        while(temp2){
            cnt++;
            temp2/=10;
        }
        if(a*temp+b*cnt>c) right=temp;
        else left=temp;
    }
    cout<<left;
    end;
}
 
/**************************************************************
    Language: C++
    Result: 正确
    Time:1 ms
    Memory:2024 kb
****************************************************************/