Codeforces 779B-Weird Rounding

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.

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.

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).

30020 3output
1input
100 9output
2input
10203049 2output
3Note
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 longint 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;}