Description:

有一个大整数.不超过1000位.假设有N位.我想删掉其中的任意S个数字.使得删除S位后,剩下位组成的数是最小的.

Input:

有多组数据数据,每组数据为两行.第一行是一个大整数.第二行是个整数S,其中S小于大整数的位数. 输入以EOF结束。

Output:

对于每组输入数据,请输出其删除后的最小数.

Sample Input:

178543

4

100002

1

Sample Output:

13

2

//#pragma comment(linker, "/STACK:16777216") //for c++ Compiler

#include <stdio.h>

#include <iostream>

#include <fstream>

#include <cstring>

#include <cmath>

#include <stack>

#include <string>

#include <map>

#include <set>

#include <list>

#include <queue>

#include <vector>

#include <algorithm>

#define Max(a,b) (((a) > (b)) ? (a) : (b))

#define Min(a,b) (((a) < (b)) ? (a) : (b))

#define Abs(x) (((x) > 0) ? (x) : (-(x)))

#define MOD 1000000007

#define pi acos(-1.0)

using namespace std;

typedef long long           ll      ;

typedef unsigned long long  ull     ;

typedef unsigned int        uint    ;

typedef unsigned char       uchar   ;

template<class T> inline void checkmin(T &a,T b){if(a>b) a=b;}

template<class T> inline void checkmax(T &a,T b){if(a<b) a=b;}

const double eps = 1e-      ;

const int N =             ;

const int M = *      ;

const ll P = 10000000097ll   ;

const int MAXN =     ;

const int INF = 0x3f3f3f3f   ;

const int offset =        ;

string ch;

char Min;

int n, m;

int main() {

    std::ios::sync_with_stdio(false);

    int i, j, t, k, u, c, v, p, numCase = ;

    while (cin >> ch) {

        n = ch.size();

        cin >> m;

        while (m--) {

            bool flag = true;//ADD

            Min = ch[];

            for (i = ; i < n; ++i) {

                if (Min > ch[i]) {

                    for (j = i; j <= n; ++j) {

                        ch[j - ] = ch[j];

                    }

                    flag = false;//ADD

                    --n;

                    break;

                } else {

                    Min = ch[i];

                }

            }

            if (flag) { //ADD

                --n;    //ADD

            }           //ADD

        }

        i = ;

        for (;;) {

            if (ch[i] == '') {

                ++i;

                continue;

            }

            break;

        }

        if (i == n) {

            cout <<  << endl;

        } else {

            for (; i < n; ++i) {

                cout << ch[i];

            }

            cout << endl;

        }

    }

    return ;

}