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 SSand 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.
Examples
input
Copy
5 3 51528
output
Copy
10028
input
Copy
3 2 102
output
Copy
100
input
Copy
1 1 1
output
Copy
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.
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <string>
#include <cstring>
#include <cstdlib>
#include <map>
#include <vector>
#include <set>
#include <queue>
#include <stack>
#include <cmath>
using namespace std;
#define ull unsigned long long
#define lli long long
#define pq priority_queue<int>
#define pql priority_queue<ll>
#define pqn priority_queue<node>
#define v vector<int>
#define vl vector<ll>
#define read(x) scanf("%d",&x)
#define lread(x) scanf("%lld",&x);
#define pt(x) printf("%d\n",(x))
#define YES printf("YES\n");
#define NO printf("NO\n");
#define gcd __gcd
#define out(x) cout<<x<<endl;
#define over cout<<endl;
#define rep(j,k) for (int i = (int)(j); i <= (int)(k); i++)
#define input(k) for (int i = 1; i <= (int)(k); i++) {scanf("%d",&a[i]) ; }
#define mem(s,t) memset(s,t,sizeof(s))
#define ok return 0;
#define TLE std::ios::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define mod(x) ((x)%9973)
#define test cout<<" ++++++ "<<endl;
//二叉树
#define lson rt<<1, l, m
#define rson rt<<1|1, m+1, r
//线段树
#define ls now<<1
#define rs now<<1|1
//int dir[6][3] = {0,0,1,0,0,-1,1,0,0,-1,0,0,0,1,0,0,-1,0};
//int dir[4][2] = {1,0,-1,0,0,1,0,-1}; //单位移动
//int dir[8][2] = {2,1,2,-1,-2,1,-2,-1,1,2,1,-2,-1,2,-1,-2};
int t,n,m,k,x,y,col,ex,ey,ans,cnt,ly;
string str;
int a[5];;
int main()
{
while(cin>>n>>k)
{
cin>>str;
if(!k)
cout<<str<<endl;
else if(n==1)
cout<<0<<endl;
else
{
if(str[0]!=‘1‘)
{str[0] = ‘1‘;k--;}
for(int i=1; i<str.size() && k; i++)
{
if(str[i]!=‘0‘)
{str[i]=‘0‘;k--;}
}
cout<<str<<endl;
}
}
}
原文地址：https://www.cnblogs.com/Shallow-dream/p/11577734.html