cf 779B Weird Rounding

题目链接：http://codeforces.com/problemset/problem/779/B

B. Weird Rounding

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard 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.

水题，比赛时没多想就交了，wa了两次，都是情况没想全

代码：

#define _CRT_SBCURE_MO_DEPRECATE
#include<iostream>
#include<stdlib.h>
#include<stdio.h>
#include<cmath>
#include<algorithm>
#include<string>
#include<string.h>
#include<set>
#include<queue>
#include<stack>
#include<functional>
using namespace std;
const int maxn = 100000000 + 10;
const int INF = 0x3f3f3f3f;

int n;
char a[maxn];
int k;

int main() {
while (scanf("%s %d", a, &k) != EOF) {
int len = strlen(a);
int sum1 = 0, sum2 = 0;//1为0的个数 2为非0个数
bool m = false;
if (k > len) {
for (int i = 0; i < len; i++) {
if (a[i] == '0') m = true;
}
if (m)printf("%d\n", len - 1);
else printf("%d\n", len);
}
else {
for (int i = len - 1; i >= 0; i--) {
if (sum1 < k) {
if (a[i] == '0')sum1++;
else sum2++;
}
}
if(sum1>=k) printf("%d\n", sum2);
else {
if (sum1 > 0) printf("%d\n", len - 1);
else printf("%d\n", len);
}
}
}

return 0;
}