A Magic Lamp HDU - 3183 （ST表）

最新推荐文章于 2021-04-29 13:15:09 发布

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Kiki在旅途中发现了一盏魔法灯，但灯里的精灵却提出了一个难题：从给定的整数中删除m位，使剩下的数字尽可能小。这涉及到算法设计和数字操作，看看如何帮助Kiki实现她的梦想。

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

Sample Output

#include <bits/stdc++.h>

using namespace std;
char a[10005];
int num[1005];
int st[10005][20];
int ans[1100];
void ST(int n)
{
    for(int i=0;i<n;i++)
        st[i][0] = i;
    for(int j = 1;(1<<j)<=n;j++)
    {
        for(int i = 0;(i+(1<<j)-1) < n;i++)
        {
            int a = st[i][j-1];
            int b = st[i+(1<<(j-1))][j-1];
            if(num[a] <= num[b]) st[i][j] = a;
            else st[i][j] = b;
        }
    }
}


int f(int l,int r)
{
    int k=(int)(log(r-l+1.0)/log(2.0));
    int b=st[l][k];
    int c=st[r-(1<<k)+1][k];
    if(num[b]<=num[c]) return b;
    else return c;
}

int main()
{
    int n,i,j,m;
    while(scanf("%s%d",a,&n) != EOF)
    {
        int len;
        len = strlen(a);
        for(i=0;i<=len;i++)
            num[i] = a[i] - '0';
        ST(len);
        m = len - n;
        i = j = 0;
        while(m--)
        {
            i = f(i,len-m-1);
            ans[j++] = num[i++];
        }
        int f = 0;
        for(i=0; i<j; i++)
        {
            if(f == 0&&ans[i] == 0)
                continue;
            printf("%d",ans[i]);
            f = 1;
        }
        if(f==0) printf("0");
        printf("\n");
    }
    return 0;
}