Weird Rounding

最新推荐文章于 2021-01-28 19:05:20 发布

tuchong5

最新推荐文章于 2021-01-28 19:05:20 发布

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本文介绍了一个编程问题的解决方案：通过最少次数的数字删除，使一个整数能被10的k次方整除。文章提供了一段C++代码示例，该程序读取一个整数n和一个指数k，然后输出需要删除的数字数量以达到目标。

Polycarp is crazy about round numbers. He especially likes the numbers divisible by10^k.

In the given number of n Polycarp wants to remove the least number of digits to get a number that is divisible by10^k. For example, ifk = 3, in the number 30020 it is enough to delete a single digit (2). In this case, the result is3000 that is divisible by 10³ = 1000.

Write a program that prints the minimum number of digits to be deleted from the given integer numbern, so that the result is divisible by 10^k. The result should not start with the unnecessary leading zero (i.e., zero can start only the number0, which is required to be written as exactly one digit).

It is guaranteed that the answer exists.

Input

The only line of the input contains two integer numbers n and k (0 ≤ n ≤ 2 000 000 000,1 ≤ k ≤ 9).

It is guaranteed that the answer exists. All numbers in the input are written in traditional notation of integers, that is, without any extra leading zeros.

Output

Print w — the required minimal number of digits to erase. After removing the appropriatew digits from the number n, the result should have a value that is divisible by 10^k. The result can start with digit0 in the single case (the result is zero and written by exactly the only digit0).

Example

Input

30020 3

Output

Input

100 9

Output

Input

10203049 2

Output

Note

In the example 2 you can remove two digits: 1 and any 0. The result is number 0 which is divisible by any number.

直接从后往前数0 就行了

#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>
#define LL long long
using namespace std;

char numb[11];

int main()
{
      int k;
      cin >> numb >> k;
      int chang = strlen(numb);
      int sum = 0, zero = 0;
      for(int i = chang - 1; i >= 0; i--)
      {
            if(zero >= k) break;
            if(numb[i] == '0')
                  zero++;
            else sum++;
      }
      if(zero == k)
            cout << sum << endl;
      else
            cout << chang - 1 << endl;


        return 0;
}