hdu 3183 A Magic Lamp (贪心 or RMQ)

A Magic Lamp

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4308    Accepted Submission(s): 1782

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 1100001 212345 254321 2

 

Sample Output

1310123321

 

Source

HDU 2009-11 Programming Contest 

这题也可以用RMQ来写，不过我暂时还不会，等会了再补起来

分析：用贪心的想法，这题不能是将所有的数字都排一个序，然后删除最大的，因为有这样的例子，1869，如果是删除9是没有删除8得到的大。

正确的应该是找到第一个比相邻的要大的位置，然后删除该位置的数。因为如果这个数不删除，它最后得到的数一定比删除的该数的数要大。

#include <iostream>#include <cstdio>#include <cstring>#define MAX_N 1005using namespace std;int main(){    char a[MAX_N];    int m;    while(cin>>a>>m)    {        int len=strlen(a);        for(int i=1;i<=m;i++)        {            int j=0;            while(j<len&&a[j]=='0') j++;            while(j<len-1&&a[j]<=a[j+1]) j++; //这里只能是a[j]>a[j+1]时才能停下来            for(int k=j;k<len;k++)                a[k]=a[k+1];        }        int flag=0;        for(int i=0;i<len-m;i++)        {            if(a[i]!='0'&&!flag)            {                flag=1;                printf("%c",a[i]);            }            else if(flag) printf("%c",a[i]);        }        if(!flag) printf("0");        printf("\n");    }    return 0;}