hdu 3183 A Magic Lamp（RMQ）

A Magic Lamp

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

Total Submission(s): 2400    Accepted Submission(s): 959


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

 

输入一个数字 删去这个数字中的m个字符 还保证剩下数字的顺序 问能组成的最小数字是什么

如果数字的总长度为len 删除m个字符还剩len-m个 则在前len-m个字符中一定有一个字符是剩下字符中的第一个

在剩下字符数一定的情况下 要保证每位上的数字尽可能小就可以了 所以要在0-len-l-m中寻找第一个字符的位置pos

找到之后 下一个字符的寻找位置从 pos+1开始 到len-2-m。。。。。

直到找到所有的数字

因为本题的过程中要找到区间最小字符的位置 所以定义一种求最小的值的方法 比较两个位置的字符大小 返回小字符的那个位置

#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string.h>
#include <string>
#include <vector>
#include <queue>

#define MEM(a,x) memset(a,x,sizeof a)
#define eps 1e-8
#define MOD 10009
#define MAXN 1010
#define MAXM 100010
#define INF 99999999
#define ll __int64
#define bug cout<<"here"<<endl
#define fread freopen("ceshi.txt","r",stdin)
#define fwrite freopen("out.txt","w",stdout)

using namespace std;

int a[MAXN],dp[MAXN][30];
char ch[1010];
int n;
char ans[1010];

int Min(int x,int y)
{
if(ch[x]<=ch[y])
return x;
return y;
}

void RMQ_init()
{
for(int i=0;i<n;i++)
dp[i][0]=i;
for(int j=1;(1<<j)<=n;j++)
for(int i=0;i+(1<<j)-1<n;i++)
dp[i][j]=Min(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);
}

int RMQ(int l,int r)
{
int k=0;
while((1<<(k+1))<=r-l+1) k++;
return Min(dp[l][k],dp[r-(1<<k)+1][k]);
}

int main()
{
// fread;
while(scanf("%s",ch)!=EOF)
{
int m;
scanf("%d",&m);
int len=strlen(ch);
n=len;
RMQ_init();
int pos=0;
int num=0;
m=len-m;
while(m--)
{
pos=RMQ(pos,len-m-1);
ans[num++]=ch[pos++];
}
int j;
for(j=0;j<num;j++)
if(ans[j]!='0')//去除先导0
break;
if(j==num)
puts("0");
else
{
for(;j<num;j++)
printf("%c",ans[j]);
puts("");
}

}
return 0;
}