codeforces 779B Weird Rounding

最新推荐文章于 2021-06-26 15:19:56 发布

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本文介绍了一个编程问题：如何通过最少次数的数字删除操作，使得一个整数能被10的K次方整除。提供了一段C++代码示例，详细解释了解决方案。

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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.

今天小赛的第三题，竟然错了12次，距结束还有3分钟时对了，我也不是很清楚，很迷。

#include<cstdio>
#include<algorithm>
using namespace std;
int main()
{
    long long n,k;
    scanf("%I64d%I64d",&n,&k);
    if(k==0)
    {
        printf("0\n");
        return 0;
    }
    int ok=0;
    int s=0,m=0,w;
    while(n>=10)
    {
        w=n%10;
        n=n/10;
        if(w==0)
            s++;
        else
            m++;
        if(s==k)
        {
            ok=1;
            break;
        }
    }
    if(ok)
        printf("%d\n",m);
    else
    {
        printf("%d\n",s+m);
    }
    return 0;
}