题意：给定一个n位的整数，要求从中去掉k位，使得剩下的数字最大。

思路：单调队列。在满足删除的数等于k 的前提下求一个不敌减的序列。

code：

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <sstream>
#include <string>
#include <vector>
#include <list>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <bitset>using namespace std;typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;const int INF=0x3fffffff;
const int inf=-INF;
const int N=100005;
const int M=2005;
const int mod=1000000007;
const double pi=acos(-1.0);#define cls(x,c) memset(x,c,sizeof(x))
#define cpy(x,a) memcpy(x,a,sizeof(a))
#define fr(i,s,n) for (int i=s;i<=n;i++)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define lrt  rt<<1
#define rrt  rt<<1|1
#define middle int m=(r+l)>>1
#define lowbit(x) (x&-x)
#define pii pair<int,int>
#define mk make_pair
#define IN freopen("in.txt","r",stdin);
#define OUT freopen("out.txt","w",stdout);int s[N];
int main()
{int n,k,x;while (~scanf("%d %d",&n,&k)&&n&&k){k=n-k;int len=0;fr(i,0,n-1){scanf("%1d",&x);//cout<<"bug"<<endl;   //每次输入一位while (len&&n-i+len>k&&x>s[len-1]) len--;  //假如还没选够k个并且当前较大，就弹出前面所选的数。其实是一个不递减的序列。if (len<k) s[len++]=x;} //cout<<len<<endl;fr(i,0,len-1) printf("%d",s[i]);puts("");}
}