【CodeForces 219B】Special Offer! Super Price 999 Bourles

最新推荐文章于 2020-08-07 14:43:10 发布

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博客介绍了一个算法问题，目标是在限定降价范围内找到以最多连续9结尾的最大价格。通过模拟方式逐步调整价格，并确保结果不超过初始价格减去最大允许降价值。

B. Special Offer! Super Price 999 Bourles!

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Polycarpus is an amateur businessman. Recently he was surprised to find out that the market for paper scissors is completely free! Without further ado, Polycarpus decided to start producing and selling such scissors.

Polycaprus calculated that the optimal celling price for such scissors would be p bourles. However, he read somewhere that customers are attracted by prices that say something like "Special Offer! Super price 999 bourles!". So Polycarpus decided to lower the price a little if it leads to the desired effect.

Polycarpus agrees to lower the price by no more than d bourles so that the number of nines at the end of the resulting price is maximum. If there are several ways to do it, he chooses the maximum possible price.

Note, Polycarpus counts only the trailing nines in a price.

Input

The first line contains two integers p and d (1 ≤ p ≤ 10¹⁸; 0 ≤ d < p) — the initial price of scissors and the maximum possible price reduction.

Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specifier.

Output

Print the required price — the maximum price that ends with the largest number of nines and that is less than p by no more than d.

The required number shouldn't have leading zeroes.

Examples

Input

1029 102

Output

Input

27191 17

Output

27189

题目大意：输出不小于 p - d 的末尾 9 最多的数，若个数相同则输出最大数

思路：模拟，依次将最后一位变为9，判断是否大于 p - d，每次用原数减去（当前位+1）* 10^位数，例如：345 - 6 = 339； 339 - 40 = 299……

这道题卡的我要哭了，wa在第一个hhh，开始 c 没初始化，结果破编译器给我自动初始化了，交上去就不行了，和本地运行结果不一样。后来又被 pow（）函数坑，pow返回值是 double 型，就算强制转换也会出现精度问题，最后自己写了快速幂，终于A了，pow不敢用了

#include<bits/stdc++.h>
typedef long long ll;
using namespace std;
ll p,d,ans,a,t,c=1,cot,k;
ll fun(ll x,ll n){ //10 的 n次幂
	ll ans = 1;
	while (n) {
		if(n&1) ans=ans*x;
		x=x*x;
		n>>= 1;
	}
	return ans;
}
int main()
{
    cin>>p>>d;
    t=p;
    while(t >= p-d && c){
        ans=t;
        k=fun(10,cot);
        c=t/k;
        a=c%10;
        if(a!=9) t-=(a+1)*k;  //已经是9则跳过
        cot++;
    }
    cout<<ans<<endl;
    return 0;
}