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

题目：

B. Weird Rounding
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
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：

给出一串长度为n的数字，问这个数字能否被10^k除尽，不能的话删除这个数字中最少几个数字能使这个数字被除尽。

**

思路：

因为题目保证结果存在
想要被10^k除尽，即统计数字n中0的个数
0的个数少于k==》数字只能删除到只剩下一个0才可能被整除
0的个数大于k==》从后往前统计删除多少个非0数字可以使连续的0的个数和k相等

**

代码：

#include<stdio.h>#include<string.h>#include<iostream>using namespace std;char n[12];int main(){    int k;    cin>>n;    scanf("%d",&k);    int i,len;    int key=0,zero=0;    len = strlen(n);    for(i=0;i<len;i++){        if(n[i]=='0'){            zero++;        }    }    if(zero<k){        key=len-1;    }else{        zero=0;        for(i=len-1;i>=0;i--){            if(zero==k){                break;            }             if(n[i]=='0'){                zero++;            }            if(n[i]!='0'){                key++;            }        }           }    cout<<key<<endl;     return 0;}