hduacm 3183 rmq

 http://acm.hdu.edu.cn/showproblem.php?pid=3183

 问题等价与取N-M个数，每次取的时候保证后面能取的个数足够，并且取的数最小  查询最小用rmq 

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>

using namespace std;


const int maxn = 10000 + 5;

int N,M,d[maxn][20];
char s[maxn];


struct RMQ{
    void init()
    {
        N = strlen(s);
        for (int i = 0;i<N;i++)
        d[i][0] = s[i];
        for (int k = 1;(1<<k)<=N;k++)
        {
            for (int i = 0;i<N;i++)
            d[i][k] = min(d[i][k-1],d[i+(1<<(k-1))][k-1]);

        }
    }
    int query(int L,int R)
    {
        int k = 0;
        while (1<<(k+1)<=R-L+1) k++;
        return min(d[L][k],d[R-(1<<k)+1][k]);
    }
};

RMQ rmq;

int main()
{
    while (~scanf("%s%d",s,&M))
    {
        rmq.init();
        M = N - M;
        int mv = 0;
        bool flag = true;
        for (int i = M;i;i--)
        {
            int c = rmq.query(mv,N-i);
            while (mv<N&&s[mv]!=c) mv++;
            mv++;
            if (c=='0'&&flag)
            continue;
            flag = false;
            printf("%c",c);

        }
        if (flag)
        printf("0");
        printf("
");
    }
    return 0;
}