Codeforces 779B-Weird Rounding

最新推荐文章于 2019-09-08 19:20:03 发布

原创最新推荐文章于 2019-09-08 19:20:03 发布 · 372 阅读

0 ·

CC 4.0 BY-SA版权

----水题同时被 2 个专栏收录

234 篇文章

订阅专栏

Codeforces

140 篇文章

订阅专栏

本文介绍了一种算法，旨在通过删除最少数量的数字使一个整数n能够被10的k次方整除。重点讨论了如何高效找到满足条件所需的最小删除次数。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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.

题意：给出数字n和数字k，要求去掉n中最少的数字，使得10^k能被得到的数整除

解题思路：若要10^k能被整除，那么这个数要么是0要么至少有k个零，又因为去掉的数字个数要尽量少，因此数字n从右往左若有大于等于k个的零，那么只要去掉从右往左数第k个零之前的所有非零数，否则就要去一直去掉数，直至只剩一个零

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <vector>
#include <set>
#include <stack>
#include <map>
#include <climits>

using namespace std;

#define LL long long

int main()
{
    LL n;
    int k,kk,nn,a[20];
    while(~scanf("%lld %d",&n,&k))
    {
        kk=k,nn=n;
        int cnt=0,sum=0,ans=0;
        while(nn) a[cnt++]=nn%10,nn/=10;
        if(n==0) cnt=1;
        for(int i=0;i<cnt;i++)
        {
            if(!a[i]) sum++;
            else ans++;
            if(sum==kk) break;
        }
        if(sum==kk) printf("%d\n",ans);
        else printf("%d\n",cnt-1);
    }
    return 0;
}