codeforces1230B水题

Ania has a large integer S. Its decimal representation has length n and doesn’t contain any leading zeroes. Ania is allowed to change at most k digits of S. She wants to do it in such a way that S 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 n and k (1≤n≤200000, 0≤k≤n) — the number of digits in the decimal representation of S and the maximum allowed number of changed digits.

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

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

Examples
Input
5 3
51528
Output
10028
Input
3 2
102
Output
100
Input
1 1
1
Output
0
Note
A number has leading zeroes if it consists of at least two digits and its first digit is 0. For example, numbers 00, 00069 and 0101 have leading zeroes, while 0, 3000 and 1010 don’t have leading zeroes.
这里要注意k可以等于0，第一次写的时候没有注意到，所以就错了，抛开这一点，其实还是很简单的

#include<iostream>
#include<stdio.h>
#include<string>
#include<cstring>
typedef long long ll;
using namespace std; 
string s;
int a[100];
int main()
{
	int n,k,j=0;
    cin>>n>>k;
   	cin>>s;
    if(n==1) {
        if(k==0)
        	cout<<s<<endl;
        else {
            cout<<0<<endl;
        }
    } else {
        if(k==0)
            cout<<s<<endl;
        else {
            if(s[0]!='1') {
            s[0]='1';
            j++;
       	 }
        for(int i=1;i<n;i++) {
            if(j==k) break;
            if(s[i]!='0') {
                s[i]='0';
                j++;
            } 
            if(j==k) break;
        }
        cout<<s<<endl;
        }
    }
    return 0;
}

这是参考别人代码又写的一个，会快一点，虽然我不太知道为什么会快。

#include<iostream>
#include<stdio.h>
#include<string>
#include<cstring>
typedef long long ll;
using namespace std; 
string s;
int a[100];
int main()
{
	int n,k,tip=1;
    cin>>n>>k;
   	cin>>s;
    if(n==1) {
        if(k==0)
        	cout<<s<<endl;
        else {
            cout<<0<<endl;
        }
    } else {
        if(k==0)
            cout<<s<<endl;
        else {
            if(s[0]!='1') {
            s[0]='1';
            k--;
       	 }
       	 /*
       	 要学习的点就在while这里，好好看，
       	 可以k值相减，这样少些判断。
       	 */
        while(tip<n && k) {
            if(s[tip]!='0') {
                k--;
                s[tip]='0';
            } 
            tip++;
        }
        cout<<s<<endl;
        }
    }
    return 0;
}