HDU-3183 A Magic Lamp （贪心）

最新推荐文章于 2021-04-29 13:15:09 发布

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本文探讨了一种算法问题，即给定一个整数和允许删除的位数，如何通过删除指定数量的位来构造出最小可能的整数。示例输入和输出展示了问题的解决过程，包括读取输入、处理字符串并输出结果。

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

123

321

#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;

int main()
{
    int m;
    char s[1002]={0};
    while(~scanf("%s%d",s,&m))
    {
        int n=strlen(s);
        int a[1002]={0};
        int b[1002]={0};
        for(int i=0;i<n;i++)
            a[i]=(int)(s[i]-'0');
        int x=0,y=m,z=0;
        m=n-m;
        while(m--)
        {
            int min=a[x];
            int index=x;
            for(int i=x;i<=y;i++)
                if(a[i]<min)
                {
                    min=a[i];
                    index=i;
                }
            b[z++]=a[index];
            x=index+1;
            y++;    
        }
        int i=0;
        while(b[i]==0&&i<z)
            i++;
        if(i==z)
            printf("0");
        else
            for(int j=i;j<z;j++)
                printf("%d",b[j]);
        printf("\n");
    }
}