RMQ hdu 3183 A Magic Lamp

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;

}