A Magic Lamp HDU

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题意就是数字串删m个字符让剩下的最小，剩下可含前导0

贪心是删掉第一个s[i]满足s[i]>s[i+1]

用rmq的话可以这样想，剩下数字串的最高位一定是s[1]~s[m+1]中的最小,设为s[x]

次高位就是s[x+1]~s[m+2]中的最小……递推一下就好

//#include<bits/stdc++.h>#include<stdio.h>#include<algorithm>#include<queue>#include<string.h>#include<iostream>#include<math.h>#include<set>#include<map>#include<vector>#include<iomanip>using namespace std;#define ll long long#define ld long double#define pb push_back#define FOR(a) for(int i=1;i<=a;i++)//const int inf=0x3f3f3f3f;#define inf 0x3f3f3f3fconst long long Linf=9e18;const int maxn=1e5+90; char s[maxn];int n,m;char num[maxn];int _min(int i,int j){return s[i]<=s[j]?i:j;}int MIN[maxn][20];void ST(){for(int i=1;i<=n;i++)MIN[i][0]=i;int k=log( (double)(n+1) )/log(2.0);for(int j=1;j<=k;j++)for(int i=1;i+(1<<j)-1<=n;i++)MIN[i][j]=_min(MIN[i][j-1],MIN[i+(1<<(j-1))][j-1]);}int rmq_min(int l,int r){int k=log((double)(r-l+1))/log(2.0);return _min(MIN[l][k],MIN[r-(1<<k)+1][k]);}int main(){while(~scanf("%s%d",s+1,&m)){int len=strlen(s+1);n=len;m=len-m;//选m个数ST();int i=1,j=1;while(m--){i=rmq_min(i,len-m);num[j++]=s[i++];}for(i=1;i<j;i++){//忽略前导0if(num[i]!='0')break;}if(i==j){puts("0");continue;//全0串}while(i<j){printf("%c",num[i]);i++;}puts("");}}

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