!= leading output chm ... cout char bsp u+

A Magic Lamp

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7170 Accepted Submission(s): 2866

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 HDU 2009-11 Programming Contest

題意就是

一個序列A[1...N]，一共N個數，除去M個數使剩下的數組成的整數最小。就是在A[1...N]中順次選取N-M個數，使值最小。

直接RMQ，找l到n-m+1的最小值，然後下一個從選取的數下一個開始，因為N-M個數，所以不能過界，要不然長度不夠。

不能有前導零，具體的代碼裏寫的。

代碼:

 1 //HDU 3183.A Magic Lamp-RMQ
 2 #include<bits/stdc++.h>
 3 using namespace std;
 4 typedef long long ll;
 5 const int maxn=1e3+10;
 6 
 7 int n,m,h;
 8 char c[maxn];
 9 int a[maxn],ans[maxn];
10 int mi[maxn][maxn];
11 
12 int Min(int x,int y)
13 {
14     return a[x]<=a[y]?x:y;
15 }
16 
17 void ST()
18 {
19     for(int i=1;i<=n;i++)
20         mi[i][0]=i;
21     for(int j=1;(1<<j)<=n;j++){
22         for(int i=1;i+(1<<j-1)<=n;i++){
23             mi[i][j]=Min(mi[i][j-1],mi[i+(1<<(j-1))][j-1]);
24         }
25     }
26 }
27 
28 int query(int l,int r)
29 {
30     int k=0;
31     while((1<<(k+1))<=r-l+1) k++;
32     int cnt=Min(mi[l][k],mi[r-(1<<k)+1][k]);
33     return cnt;
34 }
35 
36 int main()
37 {
38     while(~scanf("%s%d",c,&m)){
39         n=strlen(c);h=0;
40         for(int i=0;i<n;i++)
41             a[i+1]=c[i]-‘0‘;
42         ST();
43         int l=1;
44         m=n-m;
45         while(m>0){
46             int pos=query(l,n-m+1);
47             ans[++h]=pos;
48             l=pos+1;
49             m--;
50         }
51         if(h==0) cout<<0<<endl;
52         else{
53             int flag=0;
54             for(int i=1;i<=h;i++){
55                 if(!flag&&a[ans[i]]==0){
56                     if(i!=h) continue;
57                     else cout<<a[ans[i]];
58                 }
59                 else{
60                     flag=1;cout<<a[ans[i]];
61                 }
62             }
63             cout<<endl;
64         }
65     }
66 }

HDU 3183.A Magic Lamp-區間找最小值-RMQ(ST)