題目描述：
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?
輸入：
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.
輸出：
For each case, output the minimum result you can get in one line.
If the result contains leading zero, ignore it.
樣例輸入：
178543 4
1000001 1
100001 2
12345 2
54321 2
樣例輸出：
13
1
0
123
321
思路分析：
其實這一題就是求數列刪出後的最小值。
我們可以先用RMQ求出數列區間的最小值。
之後我們要取len-m個數，那麼我們的第一個區間邊界就是1到m+1，爲什麼呢。
我們要取len-m個數，那麼就算是後面的都取，則是len-m+1，所以區間就是[1,m+1]。
最後我們在尋找過程中用暴力就可以尋找下一個節點的左端點（所尋的數的位置）。
代碼實現：
#include<cstdio>
#include<iostream>
#include<cstring>
#include<cmath>
using namespace std;
int len,n,m,dp[1005][25],s1[1005];
char s[1005];
int Get_quary(int l,int r)
{
int k=(int)(log(double(r-l+1))/log(double(2)));
return min(dp[l][k],dp[r-(1<<k)+1][k]);
}
int main()
{
while(scanf("%s%d",s,&m)!=-1)
{
len=strlen(s);
int tot=0;
if(m<=0)
{
puts(s);
continue;
}
for(int i=0;i<len;i++)
dp[i][0]=s[i]-'0';
for(int j=1;(1<<j)<=len;j++)
for(int i=0;i+(1<<j)-1<len;i++)
dp[i][j]=min(dp[i][j-1],dp[i+(1<<j-1)][j-1]);
int l=0,r=m;
while(l<=r)
{
int p=Get_quary(l,r);
tot++;
s1[tot]=p;
int i=l;
for(;i<=r;i++)
if(p+'0'==s[i])
break;
l=i+1;
r++;
}
int i;
for(i=1;i<tot;i++)
if(s1[i]>0)
break;
if(i==tot)
{
printf("0\n");
continue;
}
for(;i<tot;i++)
printf("%d",s1[i]);
printf("\n");
memset(dp,0,sizeof(dp));
}
}