Codeforces Round #598 (Div. 3) D. Binary String Minimizing 贪心

D. Binary String Minimizing

You are given a binary string of length n (i. e. a string consisting of n characters '0' and '1').

In one move you can swap two adjacent characters of the string. What is the lexicographically minimum possible string you can obtain from the given one if you can perform no more than k moves? It is possible that you do not perform any moves at all.

Note that you can swap the same pair of adjacent characters with indices i and i+1 arbitrary (possibly, zero) number of times. Each such swap is considered a separate move.

You have to answer q independent test cases.

Input

The first line of the input contains one integer q (1≤q≤104) — the number of test cases.

The first line of the test case contains two integers n and k (1≤n≤106,1≤k≤n2) — the length of the string and the number of moves you can perform.

The second line of the test case contains one string consisting of n characters '0' and '1'.

It is guaranteed that the sum of n over all test cases does not exceed 106 (∑n≤106).

Output

For each test case, print the answer on it: the lexicographically minimum possible string of length n you can obtain from the given one if you can perform no more than k moves.

Example

input
3
8 5
11011010
7 9
1111100
7 11
1111100
output
01011110
0101111
0011111

Note

In the first example, you can change the string as follows: 110–––11010→10–––111010→011110–––10→01110–––110→0110–––1110→01011110.

In the third example, there are enough operations to make the string sorted.

题意

现在有t组数据，每组数据给你n和k；表示二进制的长度和操作次数。
每次操作你可以选择一个数和他相邻的位置进行交换，在不超过k次的操作次数情况下，使得这个字符串变得最小。

题解

贪心，每次操作最小的数，使得他尽可能的放在前面；由于里面只有0和1，其实就是操作0，把0尽可能的往前面放。

代码

#include<bits/stdc++.h>
using namespace std;

int n;
long long k;
string s;
vector<int>p;
void solve(){
	scanf("%d%lld",&n,&k);
	cin>>s;
	p.clear();
	for(int i=0;i<s.size();i++){
		if(s[i]=='0'){
			p.push_back(i);
		}
	}
	int la=-1;
	for(int i=0;i<p.size();i++){
	   // cout<<"before "<<p[i]<<" "<<k<<endl;
		if(k>p[i]-la-1){
			int cost=p[i]-la-1;
			k-=cost;
			p[i]=la+1;
			la=p[i];
		}else{
			p[i]-=k;
			break;
		}
		//cout<<"aft "<<p[i]<<" "<<k<<endl;
	}
	vector<int>o(s.size(),1);
	for(int i=0;i<p.size();i++){
		o[p[i]]=0;
	}
	for(int i=0;i<o.size();i++){
		cout<<o[i];
	}
	cout<<endl;
}
int main(){
	int t;
	scanf("%d",&t);
	while(t--)solve();
}