A Magic Lamp（HDU-3183）_amagiclanp-优快云博客

本文介绍了一个基于RMQ（Range Minimum Query）算法的解决方案，用于处理从一个给定的数字串中删除指定数量的数字，以形成可能的最小数值的问题。通过预处理数组来快速找到任意子串的最小值，该方法有效地解决了问题，同时处理了前导零的情况。

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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

题意：每组数据首先给出一个纯数字的串，然后再给出一个数字 m，现要从这个纯数串中删除 m 位，使得余下的串值最小，输出这个最小的值

思路：RMQ 找区间最小值

要注意处理前导 0，其次当两个值一样时，要输出后面的值

Source Program

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<string>
#include<cstring>
#include<cmath>
#include<ctime>
#include<algorithm>
#include<utility>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<map>
#define PI acos(-1.0)
#define E 1e-6
#define INF 0x3f3f3f3f
#define N 10001
#define LL long long
const int MOD=20091226;
const int dx[]={-1,1,0,0};
const int dy[]={0,0,-1,1};
using namespace std;
char a[N];
char res[N];
int f[N][20];
int minn(int i,int j){
    if(a[i]<=a[j])
        return i;
    return j;
}
void initMin(int n)
{
    for(int i=0;i<n;i++)
        f[i][0]=i;
    for(int j=1;(1<<j)<=n;j++)
        for(int i=0;i+(1<<j)-1<n;i++)
            f[i][j]=minn(f[i][j-1],f[i+(1<<(j-1))][j-1]);
}
int getMax(int L,int R){
    int k=0;
    while((1<<(k+1))<=R-L+1)
        k++;
    return minn(f[L][k] , f[R-(1<<k)+1][k]);
}


int main(){
    int m;
    while(scanf("%s%d",a,&m)!=EOF){
        memset(res,'\0',sizeof(res));
        int n=strlen(a);
        initMin(n);

        int len=n-m;
        int i=0,cnt=0;
        while(len--){
            i=getMax(i,n-len-1);
            res[cnt++]=a[i++];
        }

        int lead=0;//前导0
        for(int i=0;i<cnt;i++){
            lead=i;
            if(res[i]!='0')
                break;
        }

        if(lead==cnt)
            printf("0\n");
        else
            printf("%s\n",res+lead);
    }
    return 0;
}