A Magic Lamp

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 2947 Accepted Submission(s): 1149

Problem Description

Kiki likes traveling. One day she finds a magic lamp, unfortunately the genie in the lamp is not so kind. Kiki must answer a question, and then the genie will realize one of her dreams.

The question is: give you an integer, you are allowed to delete exactly m digits. The left digits will form a new integer. You should make it minimum.

You are not allowed to change the order of the digits. Now can you help Kiki to realize her dream?

Input

There are several test cases.

Each test case will contain an integer you are given (which may at most contains 1000 digits.) and the integer m (if the integer contains n digits, m will not bigger then n). The given integer will not contain leading zero.

Output

For each case, output the minimum result you can get in one line.

If the result contains leading zero, ignore it.

Sample Input

178543 4 1000001 1 100001 2 12345 2 54321 2

Sample Output

13 1 0 123 321

Source

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题意：

题意：给出一个不超过1000位的数，求删去m个数字以后形成的最小的数是多少。

题解：

贪心，前面的数要小于后面的数，用栈维护

note：

注意前导0的去除

16573675

2016-03-16 19:08:20

Accepted

15MS

1732K

1882B

C++

代码：

1 #include 
     
       2 #include 
      
        3 #include 
       
         4 #include 
        
          5 #include 
         
           6 #include 
          
            7 #include 
           
             8 #include 
            
              9 #include 
             10 #include 
              
               11 12 #define ll long long13 #define eps 1e-814 #define N 100415 #define inf 0x3ffffffffffffff16 17 using namespace std;18 19 char s[N];20 int m;21 int l;22 23 void solve()24 {25 stack
               
                 st;26 stack
                
                  ans;27 int i;28 for(i = 0;i < l;i++){29 while(!st.empty() && m > 0 && st.top()>s[i]){30 m--;31 st.pop();32 }33 if(m == 0) break;34 st.push(s[i]);35 }36 while(!st.empty() && m > 0){37 m--;38 st.pop();39 }40 //printf(" i=%d l =%d\n",i,l);41 int flag = 0;42 char te;43 while(!st.empty())44 {45 te = st.top();46 ans.push(te);47 st.pop();48 if(te != '0'){49 flag = 1;50 }51 }52 if(flag == 0){53 for(;i

转载于:https://www.cnblogs.com/njczy2010/p/5284758.html

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