RMQ hdu 3183 A Magic Lamp

//hdu 3183 A Magic Lamp
//RMQ
//用RMQ求剩下的n-m个数,第一个数肯定在第一个数和第m+1 个数之间的最小的那个数，
//包括第一和m+1，第二个数肯定在上一次求的数到第m+2 个数之间，依次类推

//注意:预处理log时，记得最大下标是所给数的长度，不是2的几次方，
//下标就是多少。求最小数时，若有多个和最小值相等的数则去最左的那个，
//要不然求下一个最小数时会忽略这些数

#include <stdio.h>
#include <string.h>
//#include <math.h>

#include <algorithm>

using namespace std;

#define comein freopen("in.txt", "r", stdin);
#define comeout freopen("out.txt", "w", stdout);
#define N 1005
#define INF 1<<30
#define eps 1e-5

char num[N];
int dp_min[20][N], pos[N];
int lo_2[N];

void RMQ(int len)
{
    for(int i = 1; i <= len; ++i)
        dp_min[0][i] = i;
    int index = lo_2[len];
    for(int i = 1; i <= index; ++i)
    {
        for(int j = 1; j + (1<<i) - 1 <= len; ++j)
        {
            if(num[dp_min[i-1][j]] <= num[dp_min[i-1][j + (1<<(i-1)) ]])
                dp_min[i][j] = dp_min[i-1][j];
            else
                dp_min[i][j] = dp_min[i-1][j + (1<<(i-1)) ];
        }
    }
}

int get_min_pos(int from, int to)
{
    int index = lo_2[to - from + 1];
    //若有多个和最小值相等的数则去最左的那个，所以这里要取等
    if(num[dp_min[index][from]] <= num[dp_min[index][to - (1<<index) + 1]])
        return dp_min[index][from];
    return dp_min[index][to - (1<<index) + 1];
}

int main()
{
    lo_2[0] = -1;

    for(int i = 1; i < N; ++i)     //求2为底的对数
        lo_2[i] = i&(i-1) ? lo_2[i-1] : lo_2[i-1] + 1;

    while(scanf("%s", &num[1]) != EOF)
    {
        int del_n, len = strlen(&num[1]);
        RMQ(len);

        scanf("%d", &del_n);
        int left = len - del_n, from = 1, to;
        to = del_n + 1;
        while(left)
        {
            pos[left-1] = get_min_pos(from, to);
            from = pos[left-1] + 1;
            to++;
            left--;
        }
        left = len - del_n;
        sort(pos, pos + left);
        int i = 0;
        while(num[pos[i]] == '0' && i < left)
            i++;
        if(i == left)
        {
            puts("0");
            continue;
        }
        for(; i < left; ++i)
            putchar(num[ pos[i] ]);
        puts("");
    }
    return 0;
}