hdu3183__A Magic Lamp(RMQ模板题)

A Magic Lamp

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

Total Submission(s): 2298 Accepted Submission(s): 909



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

Source

HDU 2009-11 Programming Contest



题意：

*************/

给出一串数字，要求删除m个数字使剩下数字按原来的顺序的组成的数最小。简单的rmq查询，这里使用的是st算法

/*************

代码如下：

#include<stdio.h>
#include<string.h>
#include<math.h>
char ch[1005], ans[1005];
int f[1005][1005];
int minn(int i, int j)
{
    return ch[i] <= ch[j] ? i : j;
}
void st(int n)
{
    int i, j;
    int k = (int)(log(double(n)) / log(2.0));
    for(i = 0; i < n; i++)
    {
        f[i][0] = i;
    }
    for(j = 1; j <= k; j++)
    {
        for(i = 0; i + (1 << j) - 1 < n; ++i)
        {
            int m = i + (1 << (j - 1));
            f[i][j] = minn(f[i][j - 1], f[m][j - 1]);
        }
    }
}
int rmq(int i, int j)		//查询区间i到j上最小值
{
    int k = (int)(log(double(j - i + 1)) / log(2.0));
    int t, m;
    m = j - (1 << k) + 1;
    return minn(f[i][k], f[m][k]);
}
int main()
{
    int m, k, n, kk;
    int maxn = 1000000;
    while(~scanf("%s%d", ch, &m))
    {
        int len = strlen(ch);
        if(len <= m)
        {
            printf("0\n");
            continue;
        }
        st(len);
        n = len - m;
        k = 0;
        for(int j = 1; j <= n; j++)	//查询len-m个最小的数存到ans中
        {
            kk = rmq(k, m + j - 1);
            k = kk + 1;
            ans[j] = ch[kk];
        }
        int flag = 0;
        for(int j = 1; j <= n; j++)
        {
            if(ans[j] != '0')
            {
                printf("%c", ans[j]);
                flag = 1;
            }
            else if(flag || j == n)
            {
                printf("%c", ans[j]);
            }
        }
        printf("\n");
    }
    return 0;
}