E. A Magic Lamp

Time Limit: 1000ms
Case Time Limit: 1000ms
Memory Limit: 32768KB
 
64-bit integer IO format: %I64d      Java class name: Main
 
 
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
 
 解题:本来要RMQ做的,不明白啊!
 
 #include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <vector>
#include <climits>
#include <algorithm>
#include <cmath>
#define LL long long
using namespace std;
char str[];
int main(){
int m,len,i,top;
while(scanf("%s %d",str,&m) == ){
len = strlen(str);
top = ;
for(i = ; i < len;){
if(m && top >= && str[top] > str[i]){m--;top--;}
else if(top != - || str[i] != '') str[++top] = str[i++];
else i++;
}
while(top >= && m){top--;m--;}
if(top == -) str[++top] = '';
str[++top] = '\0';
printf("%s\n",str);
}
return ;
}

E. A Magic Lamp的更多相关文章

  1. HDOJ 3183 A Magic Lamp

    A Magic Lamp Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Tota ...

  2. hdu 3183 A Magic Lamp(RMQ)

    题目链接:hdu 3183 A Magic Lamp 题目大意:给定一个字符串,然后最多删除K个.使得剩下的组成的数值最小. 解题思路: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,而且 ...

  3. A Magic Lamp(贪心+链表)

    A Magic Lamp Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Tota ...

  4. hdu 3183 A Magic Lamp RMQ ST 坐标最小值

    hdu 3183 A Magic Lamp RMQ ST 坐标最小值 题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3183 题目大意: 从给定的串中挑 ...

  5. HDU 3183 - A Magic Lamp - [RMQ][ST算法]

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3183 Problem DescriptionKiki likes traveling. One day ...

  6. A Magic Lamp -- hdu -- 3183

    http://acm.hdu.edu.cn/showproblem.php?pid=3183 A Magic Lamp Time Limit: 2000/1000 MS (Java/Others)   ...

  7. hdu 3183 A Magic Lamp(RMQ)

    A Magic Lamp                                                                               Time Limi ...

  8. hdu 3183 A Magic Lamp rmq或者暴力

    A Magic Lamp Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Pro ...

  9. hdu A Magic Lamp

    http://acm.hdu.edu.cn/showproblem.php?pid=3183 A Magic Lamp Time Limit: 2000/1000 MS (Java/Others)   ...

  10. HDU_3183_A Magic Lamp

    A Magic Lamp Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Tota ...

随机推荐

  1. IE如何实现text-shadow文字阴影效果呢?

    让我们头痛的是IE是不支持text-shadow效果,但为了在兼容这一问题,我们只好使用滤镜filter:shadow来处理(本人不提倡使用滤镜).filter:shadow滤镜作用与dropshad ...

  2. leetcode128 Longest Consecutive Sequence

    思路: 维护一个unordered_map,key是每个连续区间的端点,value是该区间的长度. 实现: class Solution { public: int longestConsecutiv ...

  3. viewpager的使用-新方法 5.1

    效果图: 添加依赖包: compile ‘com.android.support:design:22.2.0‘ 布局文件: <?xml version="1.0" encod ...

  4. 51nod 1283 最小周长

    一个矩形的面积为S,已知该矩形的边长都是整数,求所有满足条件的矩形中,周长的最小值.例如:S = 24,那么有{1 24} {2 12} {3 8} {4 6}这4种矩形,其中{4 6}的周长最小,为 ...

  5. C#入门(2)

    C#入门(2) Exception 基本异常的核心成员: System.Exception Property Meaning Data read-only,实现了IDirectory接口的一些键值对, ...

  6. 操作系统项目:向Linux内核添加一个系统调用

    内容: 向Linux增加一个系统调用 撰写一个应用测试程序调用该系统调用 使用ptrace或类似的工具对该测试程序进行跟踪调 环境: 1.vmware workstation 15.0.0 2.ubu ...

  7. Linux网络管理及基础设置

    一.网络管理 1 临时配置网络(ip,网关,dns) 用ifconfig命令设定网卡的IP地址: ens33网卡的IP地址为192.168.16.154, ifconfig ens33 192.168 ...

  8. Bootstrap响应式布局(1)

    <!DOCTYPE html><html><head><meta http-equiv="Content-Type" content=&q ...

  9. java在线聊天项目0.1版本 制作客户端窗体,使用swing(用户界面开发工具包)和awt(抽象窗口工具包)

    建立Chat项目,并在项目中创建窗口类 package com.swift; import java.awt.BorderLayout; import javax.swing.JFrame; impo ...

  10. ios多线程原理及runloop介绍

    一.线程概述 有些程序是一条直线,起点到终点:有些程序是一个圆,不断循环,直到将它切断.直线的如简单的Hello World,运行打印完,它的生命周期便结束了,像昙花一现那样:圆如操作系统,一直运行直 ...