B. Weird Rounding - Codeforces Round #402 (Div. 2)

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本文介绍了一个算法问题：如何从一个整数中删除最少数量的数字，使其成为可被10^k整除的圆数。文章提供了一个C语言实现的示例程序，通过解析输入的数字和指定的指数k，输出最小的删除次数。

Polycarp is crazy about round numbers. He especially likes the numbers divisible by 10k.

In the given number of n Polycarp wants to remove the least number of digits to get a number that is divisible by 10k. For example, if k = 3, in the number 30020 it is enough to delete a single digit (2). In this case, the result is 3000 that is divisible by 103 = 1000.

Write a program that prints the minimum number of digits to be deleted from the given integer number n, so that the result is divisible by 10k. The result should not start with the unnecessary leading zero (i.e., zero can start only the number 0, 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 appropriate w digits from the number n, the result should have a value that is divisible by 10k. The result can start with digit 0 in the single case (the result is zero and written by exactly the only digit 0).

Examples
input
30020 3
output
1
input
100 9
output
2
input
10203049 2
output
3
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.

#include<stdio.h>
#include<string.h>
int main()
{
    char str[100];int k;
    scanf("%s%d",str,&k);
    int len=strlen(str);

    int i=len-1,ans=0;
    for(;k>0&&i>=0;i--){
        if(str[i]=='0') k--;
        else ans++;
    }
    if(k>0){
        bool f=false;
        for(int i=0;i<len;i++)
            if(str[i]=='0')
                f=true;
        printf("%d\n",f?len-1:len);
    }
    else printf("%d\n",ans);

    return 0;
}