Description 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 allowe…

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,而且取的数最小,查询最小的操作用RMQ优化. #include <cstdio> #include <cstring> #include <algorithm> using namespace std; const int maxn = 10005; int N, M, d[m…

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: 从给定的串中挑出来m个数使得剩余的数字最小,串的序列不能改变 思路: 将问题转化为求在n个数中挑选n-m个数,使之最小.假设最极端的情况,所有最大的数字都在左侧,占据了m个位置,那么我们需要挑选的最小的数字的第一位就是在m+1位上.依次类推,第二位在m+2位上,最后一位也就在原串的最后一位上.反过来,假设最大的数…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3183 Problem DescriptionKiki 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 d…

A Magic Lamp Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Problem Description 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…

A Magic Lamp                                                                               Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Problem Description Kiki likes traveling. One day she finds a magic lamp, u…

原题目 A Magic Lamp Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3964    Accepted Submission(s): 1605 Problem Description Kiki likes traveling. One day she finds a magic lamp, unfortunately the…

直接模拟   如果后一位比前一位小,那就一直 向前 pop()掉 维护他单调递增: #include<iostream> #include<cstring> #include<algorithm> #include<cmath> #include<stdio.h> #include<vector> using namespace std; ]; ],num; int main( ) { while( scanf("%s%d&…

第一种做法是贪心做法,只要前面的数比后面的大就把他删掉,这种做法是正确的,也比较好理解,这里就不说了,我比较想说一下ST算法,RMQ的应用 主要是返回数组的下标,RMQ要改成<=(这里是个坑点,取连续数是可以的),他的转移方程为x = dp[i-1][j],y = dp[i-1][j+1<<(i-1)]; dp[i][j] = a[x] <= a[y] ? x : y; 这里既是转化为取n-m个数,首先在1 ~ m+1中必然选一个数,所以这个数就是n-m数中第一个数,记录所选的位置…

<题目链接> <转载于 >>>  > 题目大意: 给出一个长度不超过1000位的数,求删去m位数字以后形成的最小的数字是多少. 解题分析: 分析:我们可以把题目转化为这样一个模型:从A[1].A[2].…….A[n] n个数中选出n-m个数,使得组成的数最小.一.使用RMQ,设原数字长为n,那么除去m个数字后还剩n-m个数字.    下面的(一).(二)用到了抽屉原理(1)因为有n-m个数字,那么在0到m-1位置中最小的那个数字必是结果中的第一个数字,记录其位置为…

1.题目大意是,给你一个1000位的数,要你删掉m个为,求结果最小数. 思路:在n个位里面删除m个位.也就是找出n-m个位组成最小数 所以在区间 [0, m]里面找最小的数.相应的下标标号i 接着找区间 [i+1,m++]里面的最小数.对于下标为ii 接着找区间 [ii+1,m++]里面的最小数-- 这样就会找n-m个数了.区间这样安排的目的是为了保证取出来的数的顺序. 2代码: #include<cstdio> #include<cstring> #include<cmat…

#include <stdio.h> #include <string.h> #include <iostream> #include <algorithm> #include <vector> #include <queue> #include <stack> #include <set> #include <map> #include <string> #include <ma…

http://acm.hdu.edu.cn/showproblem.php?pid=3183 A Magic Lamp Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2515    Accepted Submission(s): 983 Problem Description Kiki likes traveling. One day…

http://acm.hdu.edu.cn/showproblem.php?pid=3183 A Magic Lamp Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2556    Accepted Submission(s): 999 Problem Description Kiki likes traveling. One day…

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…

A Magic Lamp Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 989    Accepted Submission(s): 371 Problem Description Kiki likes traveling. One day she finds a magic lamp, unfortunately the genie…

A Magic Lamp Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4694    Accepted Submission(s): 1947 Problem Description Kiki likes traveling. One day she finds a magic lamp, unfortunately the geni…

A Magic Lamp Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 7170    Accepted Submission(s): 2866 Problem Description Kiki likes traveling. One day she finds a magic lamp, unfortunately the geni…

A Magic Lamp Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2584    Accepted Submission(s): 1010 Problem Description Kiki likes traveling. One day she finds a magic lamp, unfortunately the gen…

A Magic Lamp Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2521    Accepted Submission(s): 986 Problem Description Kiki likes traveling. One day she finds a magic lamp, unfortunately the genie…

HDU 4421 Bit Magic pid=4421" target="_blank" style="">题目链接 题意:就依据题目,给定b数组.看能不能构造出一个符合的a数组 思路:把每一个数字的每一个二进制位单独考虑.就变成一个2-sat题目了,依据题目中的式子建立2-sat的边.然后每一位跑2-sat.假设每位都符合.就是YES,假设有一位不符合就是NO 代码: #include <cstdio> #include <cst…

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…

A Magic Lamp Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2937    Accepted Submission(s): 1145 Problem Description Kiki likes traveling. One day she finds a magic lamp, unfortunately the gen…

HDU 6187 Destroy Walls (思维,最大生成树) Destroy Walls *Time Limit: 8000/4000 MS (Java/Others) Memory Limit: 132768/132768 K (Java/Others) Total Submission(s): 1784 Accepted Submission(s): 692 * Problem Description Long times ago, there are beautiful histor…

http://acm.hdu.edu.cn/showproblem.php?pid=3183 题意:给出一个数,可以删除掉其中m个字符,要使得最后的数字最小,输出最后的数字(忽略前导零). 思路:设数的长度为n,那么我们要删除其中m个最大的字符,可以转化为我们保留其中的n-m个最小的字符.对于第一个数,它存在的区间必定在[1,m+1]里面,因为我们要保证后面[m+2,n]区间有n-m-1个字符.当找到第一个数的下标为tmp的时候,第二个数的区间就是[tmp+1,m+2]……以此类推,直到找到最后…

原文地址:https://blog.csdn.net/acdreamers/article/details/8692384 题意: 对于一个序列A[1...N],一共N个数,除去M个数使剩下的数组成的整数最小. 也就是说在A[1...N]中顺次选取N-M个数,使值最小. 本题很有技巧性,一开始我总是想不明白,后来在纸上画了一下,大概明白了是怎么回事. 它主要是基于以下事实: 对于序列A[1...N],选取N-M个数,使组成的值最小,而且顺序不能交换,既然要选取N-M个,那么可以容易知道这N-M位…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3182 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Problem Description In the mysterious forest, there is a group of Magi. Most of them like to eat human beings, so the…

题目链接:pid=3182">http://acm.hdu.edu.cn/showproblem.php?pid=3182 Problem Description In the mysterious forest, there is a group of Magi. Most of them like to eat human beings, so they are called "The Ogre Magi", but there is an special one…

Bit Magic Problem Description Yesterday, my teacher taught me about bit operators: and (&), or (|), xor (^). I generated a number table a[N], and wrote a program to calculate the matrix table b[N][N] using three kinds of bit operator. I thought my ac…

Bit Magic Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3040    Accepted Submission(s): 871 Problem Description Yesterday, my teacher taught me about bit operators: and (&), or (|), xor (^). I…