Ania and Minimizing

2024-06-07 07:22:43

Ania has a large integer SS. Its decimal representation has length nn and doesn’t contain any leading zeroes. Ania is allowed to change at most kk digits of SS. She wants to do it in such a way that SS still won’t contain any leading zeroes and it’ll be minimal possible. What integer will Ania finish with?

Input

The first line contains two integers nn and kk (1≤n≤2000001≤n≤200000, 0≤k≤n0≤k≤n) — the number of digits in the decimal representation of SS and the maximum allowed number of changed digits.

The second line contains the integer SS. It’s guaranteed that SS has exactly nn digits and doesn’t contain any leading zeroes.

Output

Output the minimal possible value of SS which Ania can end with. Note that the resulting integer should also have nn digits.

Sample Input

5 3
51528

3 2
102

1 1
1

Sample Output

10028

100

0

Note

A number has leading zeroes if it consists of at least two digits and its first digit is 00. For example, numbers 0000, 0006900069 and 01010101 have leading zeroes, while 00, 30003000 and 10101010 don’t have leading zeroes.

题目大意:

一个 n 位数，经过 k 次变化，输出变化 k 次后的最小值，且这个 n 位数没有前导 0。

解题思路：

如果是多位数使首位为1，其余位为0，如果为单位数直接为0。

代码如下：

#include <stdio.h>
#include <string.h>
char s[200002];
int main()
{int n,k,i=0,j=0;scanf("%d%d%s",&n,&k,&s);for(i=0;i<n;i++){if(j<k&&k>0){if(s[0]!='1'&&i==0){s[i]='1';j++;}else if(i!=0&&s[i]!='0'){s[i]='0';j++;}if (n==1){s[i]='0';}}else{break;}}printf("%s",s);
}

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